ChemistryQ1–66 of 66 questions
Page 1 of 1 · English
1
ChemistryEasyMCQAIPMT · 1994
In the final answer of the expression $\frac{(29.2 - 20.2)(1.79 \times 10^5)}{1.37}$,the number of significant figures is
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(B) Step $1$: Perform the subtraction in the numerator: $(29.2 - 20.2) = 9.0$.
Since the subtraction rule states that the result should have the same number of decimal places as the term with the fewest decimal places,$9.0$ has $1$ decimal place (two significant figures).
Step $2$: The expression becomes $\frac{9.0 \times 1.79 \times 10^5}{1.37}$.
Step $3$: According to the multiplication and division rule,the final result should have the same number of significant figures as the term with the fewest significant figures.
Here,$9.0$ has $2$ significant figures,$1.79$ has $3$ significant figures,and $1.37$ has $3$ significant figures.
Therefore,the final answer must have $2$ significant figures.
Since the subtraction rule states that the result should have the same number of decimal places as the term with the fewest decimal places,$9.0$ has $1$ decimal place (two significant figures).
Step $2$: The expression becomes $\frac{9.0 \times 1.79 \times 10^5}{1.37}$.
Step $3$: According to the multiplication and division rule,the final result should have the same number of significant figures as the term with the fewest significant figures.
Here,$9.0$ has $2$ significant figures,$1.79$ has $3$ significant figures,and $1.37$ has $3$ significant figures.
Therefore,the final answer must have $2$ significant figures.
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2
ChemistryEasyMCQAIPMT · 1994
Which one of the following is not isoelectronic with ${O^{2-}}$?
A
${N^{3-}}$
B
${F^{-}}$
C
${Tl^{+}}$
D
${Na^{+}}$
Solution
(C) Isoelectronic species are those that have the same number of electrons.
The oxygen anion ${O^{2-}}$ has $8 + 2 = 10$ electrons.
Checking the options:
$1$. ${N^{3-}}$ has $7 + 3 = 10$ electrons.
$2$. ${F^{-}}$ has $9 + 1 = 10$ electrons.
$3$. ${Na^{+}}$ has $11 - 1 = 10$ electrons.
$4$. ${Tl^{+}}$ (Thallium ion) has an atomic number of $81$,so ${Tl^{+}}$ has $81 - 1 = 80$ electrons.
Therefore,${Tl^{+}}$ is not isoelectronic with ${O^{2-}}$.
The oxygen anion ${O^{2-}}$ has $8 + 2 = 10$ electrons.
Checking the options:
$1$. ${N^{3-}}$ has $7 + 3 = 10$ electrons.
$2$. ${F^{-}}$ has $9 + 1 = 10$ electrons.
$3$. ${Na^{+}}$ has $11 - 1 = 10$ electrons.
$4$. ${Tl^{+}}$ (Thallium ion) has an atomic number of $81$,so ${Tl^{+}}$ has $81 - 1 = 80$ electrons.
Therefore,${Tl^{+}}$ is not isoelectronic with ${O^{2-}}$.
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3
ChemistryEasyMCQAIPMT · 1994
For which of the following sets of four quantum numbers,an electron will have the highest energy?
$n$ $l$ $m$ $s$
$n$ $l$ $m$ $s$
A
$n=3, l=2, m=1, s=+1/2$
B
$n=5, l=0, m=0, s=-1/2$
C
$n=4, l=1, m=0, s=-1/2$
D
$n=4, l=2, m=-1, s=+1/2$
Solution
(D) The energy of an orbital is determined by the $(n+l)$ rule.
Calculating $(n+l)$ values for each set:
$A: 3+2=5$
$B: 5+0=5$
$C: 4+1=5$
$D: 4+2=6$
Since orbital $D$ has the highest $(n+l)$ value of $6$,it has the highest energy.
Calculating $(n+l)$ values for each set:
$A: 3+2=5$
$B: 5+0=5$
$C: 4+1=5$
$D: 4+2=6$
Since orbital $D$ has the highest $(n+l)$ value of $6$,it has the highest energy.
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4
ChemistryMediumMCQAIPMT · 1994
The total number of valence electrons in $4.2 \ g$ of $N_3^-$ ion is ............. $N_A$ ($N_A$ is the Avogadro's number).
A
$1.6$
B
$3.2$
C
$2.1$
D
$4.2$
Solution
(A) The molar mass of $N_3^-$ ion is $(3 \times 14) = 42 \ g/mol$.
Number of moles of $N_3^-$ in $4.2 \ g = \frac{4.2 \ g}{42 \ g/mol} = 0.1 \ mol$.
Each $N_3^-$ ion has $(3 \times 5) + 1 = 16$ valence electrons.
Total valence electrons $= 0.1 \ mol \times 16 \times N_A = 1.6 \ N_A$.
Number of moles of $N_3^-$ in $4.2 \ g = \frac{4.2 \ g}{42 \ g/mol} = 0.1 \ mol$.
Each $N_3^-$ ion has $(3 \times 5) + 1 = 16$ valence electrons.
Total valence electrons $= 0.1 \ mol \times 16 \times N_A = 1.6 \ N_A$.
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5
ChemistryMediumMCQAIPMT · 1994
The table shown below gives the bond dissociation energies $E_{diss}$ for single covalent bonds of carbon $C$ atoms with elements $A, B, C$ and $D.$ Which element has the smallest atoms?
| $Bond$ | $E_{diss} (kJ \ mol^{-1})$ |
|---|---|
| $C-A$ | $240$ |
| $C-B$ | $328$ |
| $C-C$ | $276$ |
| $C-D$ | $485$ |
A
$A$
B
$B$
C
$C$
D
$D$
Solution
(D) Bond dissociation energy is inversely proportional to the bond length,which in turn depends on the atomic size of the bonded atoms.
Smaller atoms form shorter,stronger bonds,resulting in higher bond dissociation energy.
Comparing the given values: $485 \ kJ \ mol^{-1} > 328 \ kJ \ mol^{-1} > 276 \ kJ \ mol^{-1} > 240 \ kJ \ mol^{-1}$.
Since element $D$ has the highest bond dissociation energy $(485 \ kJ \ mol^{-1})$,it must have the smallest atomic size among the given elements.
Smaller atoms form shorter,stronger bonds,resulting in higher bond dissociation energy.
Comparing the given values: $485 \ kJ \ mol^{-1} > 328 \ kJ \ mol^{-1} > 276 \ kJ \ mol^{-1} > 240 \ kJ \ mol^{-1}$.
Since element $D$ has the highest bond dissociation energy $(485 \ kJ \ mol^{-1})$,it must have the smallest atomic size among the given elements.
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6
ChemistryDifficultMCQAIPMT · 1994
Which of the following molecules does not possess a permanent dipole moment?
A
$H_2S$
B
$SO_2$
C
$CS_2$
D
$SO_3$
Solution
(C) molecule possesses a permanent dipole moment if it has a non-zero net dipole moment due to its geometry and electronegativity difference.
$H_2S$ has a bent geometry and a net dipole moment.
$SO_2$ has a bent geometry and a net dipole moment.
$CS_2$ has a linear geometry $(S=C=S)$,where the two $C=S$ bond dipoles are equal in magnitude and opposite in direction,canceling each other out,resulting in a net dipole moment of $0$.
$SO_3$ is a planar trigonal molecule with a net dipole moment of $0$,but among the given options,$CS_2$ is a classic example of a linear non-polar molecule.
$H_2S$ has a bent geometry and a net dipole moment.
$SO_2$ has a bent geometry and a net dipole moment.
$CS_2$ has a linear geometry $(S=C=S)$,where the two $C=S$ bond dipoles are equal in magnitude and opposite in direction,canceling each other out,resulting in a net dipole moment of $0$.
$SO_3$ is a planar trigonal molecule with a net dipole moment of $0$,but among the given options,$CS_2$ is a classic example of a linear non-polar molecule.
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7
ChemistryMediumMCQAIPMT · 1994
The bond order is maximum in
A
$O_2$
B
$O_2^{-}$
C
$O_2^{+}$
D
$O_2^{2-}$
Solution
(C) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$.
For $O_2$ ($16$ electrons): $B.O. = \frac{1}{2} (10 - 6) = 2.0$.
For $O_2^{-}$ ($17$ electrons): $B.O. = \frac{1}{2} (10 - 7) = 1.5$.
For $O_2^{+}$ ($15$ electrons): $B.O. = \frac{1}{2} (10 - 5) = 2.5$.
For $O_2^{2-}$ ($18$ electrons): $B.O. = \frac{1}{2} (10 - 8) = 1.0$.
Comparing these values,the bond order is maximum in $O_2^{+}$.
For $O_2$ ($16$ electrons): $B.O. = \frac{1}{2} (10 - 6) = 2.0$.
For $O_2^{-}$ ($17$ electrons): $B.O. = \frac{1}{2} (10 - 7) = 1.5$.
For $O_2^{+}$ ($15$ electrons): $B.O. = \frac{1}{2} (10 - 5) = 2.5$.
For $O_2^{2-}$ ($18$ electrons): $B.O. = \frac{1}{2} (10 - 8) = 1.0$.
Comparing these values,the bond order is maximum in $O_2^{+}$.
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8
ChemistryMediumMCQAIPMT · 1994
The temperature of the gas is raised from $27\,^oC$ to $927\,^oC$. The root mean square velocity is:
A
$\sqrt{927/27}$ times the earlier value
B
Same as before
C
Halved
D
Doubled
Solution
(D) The root mean square velocity $(U_{rms})$ is given by the formula $U_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $U_{rms} \propto \sqrt{T}$,the ratio of velocities is $\frac{U_2}{U_1} = \sqrt{\frac{T_2}{T_1}}$.
Given $T_1 = 27 + 273 = 300 \, K$ and $T_2 = 927 + 273 = 1200 \, K$.
Substituting the values: $\frac{U_2}{U_1} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
Therefore,$U_2 = 2 \times U_1$,which means the velocity is doubled.
Since $U_{rms} \propto \sqrt{T}$,the ratio of velocities is $\frac{U_2}{U_1} = \sqrt{\frac{T_2}{T_1}}$.
Given $T_1 = 27 + 273 = 300 \, K$ and $T_2 = 927 + 273 = 1200 \, K$.
Substituting the values: $\frac{U_2}{U_1} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
Therefore,$U_2 = 2 \times U_1$,which means the velocity is doubled.
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9
ChemistryMediumMCQAIPMT · 1994
$50 \ mL$ of hydrogen diffuses out through a small hole from a vessel in $20 \ min$. The time needed for $40 \ mL$ of oxygen to diffuse out is ............. $min$.
A
$12$
B
$64$
C
$8$
D
$32$
Solution
(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
For two gases,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given $r_{H_2} = \frac{50 \ mL}{20 \ min} = 2.5 \ mL/min$ and $M_{H_2} = 2 \ g/mol$.
For oxygen,$M_{O_2} = 32 \ g/mol$ and volume $V = 40 \ mL$.
Let the time taken be $t$. Then $r_{O_2} = \frac{40}{t}$.
Substituting the values: $\frac{2.5}{40/t} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
$\frac{2.5 \times t}{40} = 4 \implies 2.5 \times t = 160$.
$t = \frac{160}{2.5} = 64 \ min$.
For two gases,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given $r_{H_2} = \frac{50 \ mL}{20 \ min} = 2.5 \ mL/min$ and $M_{H_2} = 2 \ g/mol$.
For oxygen,$M_{O_2} = 32 \ g/mol$ and volume $V = 40 \ mL$.
Let the time taken be $t$. Then $r_{O_2} = \frac{40}{t}$.
Substituting the values: $\frac{2.5}{40/t} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
$\frac{2.5 \times t}{40} = 4 \implies 2.5 \times t = 160$.
$t = \frac{160}{2.5} = 64 \ min$.
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10
ChemistryDifficultMCQAIPMT · 1994
Which one of the following is most soluble?
A
$CuS$ $(K_{sp} = 8 \times 10^{-37})$
B
$MnS$ $(K_{sp} = 7 \times 10^{-16})$
C
$Bi_2S_3$ $(K_{sp} = 1 \times 10^{-70})$
D
$Ag_2S$ $(K_{sp} = 6 \times 10^{-51})$
Solution
(B) The solubility $(S)$ of a salt is related to its solubility product constant $(K_{sp})$.
For salts of the same type (e.g.,$AB$ type),solubility is directly proportional to the $K_{sp}$.
Comparing the given values: $MnS$ has the largest $K_{sp}$ value $(7 \times 10^{-16})$,which indicates it is the most soluble among the given options.
For salts of the same type (e.g.,$AB$ type),solubility is directly proportional to the $K_{sp}$.
Comparing the given values: $MnS$ has the largest $K_{sp}$ value $(7 \times 10^{-16})$,which indicates it is the most soluble among the given options.
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11
ChemistryMediumMCQAIPMT · 1994
At $80\,^{\circ}C,$ distilled water has $[H_3O^{+}]$ concentration equal to $1 \times 10^{-6} \, mol/L.$ The value of $K_w$ at this temperature will be
A
$1 \times 10^{-6}$
B
$1 \times 10^{-9}$
C
$1 \times 10^{-12}$
D
$1 \times 10^{-15}$
Solution
(C) In distilled water,the auto-ionization reaction is $2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq).$
Since the water is neutral,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}] = 1 \times 10^{-6} \, mol/L.$
The ionic product of water,$K_w,$ is defined as $K_w = [H_3O^{+}][OH^{-}].$
Substituting the given values: $K_w = (1 \times 10^{-6}) \times (1 \times 10^{-6}) = 1 \times 10^{-12}.$
Therefore,the correct option is $C.$
Since the water is neutral,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}] = 1 \times 10^{-6} \, mol/L.$
The ionic product of water,$K_w,$ is defined as $K_w = [H_3O^{+}][OH^{-}].$
Substituting the given values: $K_w = (1 \times 10^{-6}) \times (1 \times 10^{-6}) = 1 \times 10^{-12}.$
Therefore,the correct option is $C.$
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12
ChemistryMediumMCQAIPMT · 1994
For the following combustion reaction occurring in an automobile: $2C_{8}H_{18(g)} + 25O_{2(g)} \to 16CO_{2(g)} + 18H_{2}O_{(g)}$,what are the signs of $\Delta H$,$\Delta S$,and $\Delta G$?
A
$+, -, +$
B
$-, +, -$
C
$-, +, +$
D
$+, +, -$
Solution
(B) The given reaction is a combustion reaction,which is exothermic,so $\Delta H < 0$ (negative).
For the change in the number of moles of gas,$\Delta n_g = (16 + 18) - (2 + 25) = 34 - 27 = +7$.
Since $\Delta n_g > 0$,the entropy increases,so $\Delta S > 0$ (positive).
Combustion reactions in automobiles are spontaneous processes,therefore $\Delta G < 0$ (negative).
Thus,the signs are $\Delta H = -$,$\Delta S = +$,and $\Delta G = -$,which corresponds to option $B$.
For the change in the number of moles of gas,$\Delta n_g = (16 + 18) - (2 + 25) = 34 - 27 = +7$.
Since $\Delta n_g > 0$,the entropy increases,so $\Delta S > 0$ (positive).
Combustion reactions in automobiles are spontaneous processes,therefore $\Delta G < 0$ (negative).
Thus,the signs are $\Delta H = -$,$\Delta S = +$,and $\Delta G = -$,which corresponds to option $B$.
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13
ChemistryMediumMCQAIPMT · 1994
Which substance is serving as a reducing agent in the following reaction: $14H^{+} + Cr_2O_7^{2-} + 3Ni \to 2Cr^{3+} + 7H_2O + 3Ni^{2+}$?
A
$H_2O$
B
$Ni$
C
$H^{+}$
D
$Cr_2O_7^{2-}$
Solution
(B) In the given reaction: $14H^{+} + Cr_2O_7^{2-} + 3Ni \to 2Cr^{3+} + 7H_2O + 3Ni^{2+}$
$1$. The oxidation state of $Ni$ increases from $0$ (in elemental form) to $+2$ (in $Ni^{2+}$ ion).
$2$. Since $Ni$ undergoes oxidation (loss of electrons),it acts as the reducing agent.
$3$. The oxidation state of $Cr$ in $Cr_2O_7^{2-}$ decreases from $+6$ to $+3$,meaning $Cr_2O_7^{2-}$ acts as the oxidizing agent.
Therefore,the correct option is $(B)$.
$1$. The oxidation state of $Ni$ increases from $0$ (in elemental form) to $+2$ (in $Ni^{2+}$ ion).
$2$. Since $Ni$ undergoes oxidation (loss of electrons),it acts as the reducing agent.
$3$. The oxidation state of $Cr$ in $Cr_2O_7^{2-}$ decreases from $+6$ to $+3$,meaning $Cr_2O_7^{2-}$ acts as the oxidizing agent.
Therefore,the correct option is $(B)$.
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14
ChemistryEasyMCQAIPMT · 1994
The oxidation state of $I$ in $H_4IO_6^-$ is
A
$+7$
B
$+5$
C
$+1$
D
$-1$
Solution
(A) Let the oxidation state of $I$ be $x$.
In the ion $H_4IO_6^-$,the sum of the oxidation states of all atoms equals the charge on the ion.
$4(+1) + x + 6(-2) = -1$
$4 + x - 12 = -1$
$x - 8 = -1$
$x = +7$
Therefore,the oxidation state of $I$ is $+7$.
In the ion $H_4IO_6^-$,the sum of the oxidation states of all atoms equals the charge on the ion.
$4(+1) + x + 6(-2) = -1$
$4 + x - 12 = -1$
$x - 8 = -1$
$x = +7$
Therefore,the oxidation state of $I$ is $+7$.
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15
ChemistryEasyMCQAIPMT · 1994
The formula for calcium chlorite is
A
$Ca(ClO_4)_2$
B
$Ca(ClO_3)_2$
C
$CaCl(O_2)$
D
$Ca(ClO_2)_2$
Solution
(D) The chemical formula for calcium chlorite is derived from the calcium ion $(Ca^{2+})$ and the chlorite ion $(ClO_2^-)$.
To balance the charges,two chlorite ions are required for every one calcium ion,resulting in the formula $Ca(ClO_2)_2$.
Calcium chlorite is a white granular solid.
To balance the charges,two chlorite ions are required for every one calcium ion,resulting in the formula $Ca(ClO_2)_2$.
Calcium chlorite is a white granular solid.
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16
ChemistryMediumMCQAIPMT · 1994
Which pair of substances gives the same gaseous product when they react with water?
A
$Ca$ and $CaH_2$
B
$Na$ and $Na_2O_2$
C
$K$ and $KO_2$
D
$Ba$ and $BaO_2$
Solution
(A) For option $A$:
$Ca + 2H_2O \to Ca(OH)_2 + H_2(g)$
$CaH_2 + 2H_2O \to Ca(OH)_2 + 2H_2(g)$
Both reactions produce hydrogen gas $(H_2)$.
For option $B$:
$2Na + 2H_2O \to 2NaOH + H_2(g)$
$2Na_2O_2 + 2H_2O \to 4NaOH + O_2(g)$
These produce different gases ($H_2$ and $O_2$).
For option $C$:
$2K + 2H_2O \to 2KOH + H_2(g)$
$2KO_2 + 2H_2O \to 2KOH + H_2O_2 + O_2(g)$
These produce different gases.
For option $D$:
$Ba + 2H_2O \to Ba(OH)_2 + H_2(g)$
$BaO_2 + 2H_2O \to Ba(OH)_2 + H_2O_2$
These do not produce the same gaseous product.
$Ca + 2H_2O \to Ca(OH)_2 + H_2(g)$
$CaH_2 + 2H_2O \to Ca(OH)_2 + 2H_2(g)$
Both reactions produce hydrogen gas $(H_2)$.
For option $B$:
$2Na + 2H_2O \to 2NaOH + H_2(g)$
$2Na_2O_2 + 2H_2O \to 4NaOH + O_2(g)$
These produce different gases ($H_2$ and $O_2$).
For option $C$:
$2K + 2H_2O \to 2KOH + H_2(g)$
$2KO_2 + 2H_2O \to 2KOH + H_2O_2 + O_2(g)$
These produce different gases.
For option $D$:
$Ba + 2H_2O \to Ba(OH)_2 + H_2(g)$
$BaO_2 + 2H_2O \to Ba(OH)_2 + H_2O_2$
These do not produce the same gaseous product.
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17
ChemistryMediumMCQAIPMT · 1994
Which of the following statements about $H_3BO_3$ is not correct?
A
It is a strong tribasic acid
B
It is prepared by acidifying an aqueous solution of borax
C
It has a layer structure in which planar $BO_3$ units are joined by hydrogen bonds
D
It does not act as proton donor but acts as a Lewis acid by accepting hydroxyl ion
Solution
(A) $H_3BO_3$ (orthoboric acid) is a very weak monoprotic Lewis acid,not a strong tribasic acid.
It acts as a Lewis acid by accepting an $OH^-$ ion from water molecules,releasing $H^+$ ions in the process: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$.
Therefore,the statement that it is a strong tribasic acid is incorrect.
It acts as a Lewis acid by accepting an $OH^-$ ion from water molecules,releasing $H^+$ ions in the process: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$.
Therefore,the statement that it is a strong tribasic acid is incorrect.
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18
ChemistryMediumMCQAIPMT · 1994
Carbon and silicon belong to group $IV$. The maximum coordination number of carbon in commonly occurring compounds is $4$,whereas that of silicon is $6$. This is due to
A
Large size of silicon
B
More electropositive nature of silicon
C
Availability of low lying $d$-orbitals in silicon
D
Both $(a)$ and $(b)$
Solution
(C) Carbon $(C)$ has the electronic configuration $2s^2 2p^2$ and lacks $d$-orbitals in its valence shell,limiting its covalency to $4$.
Silicon $(Si)$ has the electronic configuration $3s^2 3p^2$ and possesses vacant $3d$-orbitals.
Due to the availability of these low-lying vacant $d$-orbitals,silicon can expand its octet and accommodate more ligands,allowing it to exhibit a coordination number of $6$ (e.g.,in $[SiF_6]^{2-}$).
Silicon $(Si)$ has the electronic configuration $3s^2 3p^2$ and possesses vacant $3d$-orbitals.
Due to the availability of these low-lying vacant $d$-orbitals,silicon can expand its octet and accommodate more ligands,allowing it to exhibit a coordination number of $6$ (e.g.,in $[SiF_6]^{2-}$).
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19
ChemistryMediumMCQAIPMT · 1994
Lassaigne's test for the detection of nitrogen fails in
A
$NH_2CONHNH_2 \cdot HCl$
B
$NH_2NH_2 \cdot HCl$
C
$NH_2CONH_2$
D
$C_6H_5NHNH_2 \cdot HCl$
Solution
(B) Lassaigne's test requires the presence of both carbon and nitrogen in an organic compound to form sodium cyanide $(NaCN)$ upon fusion with sodium metal.
Compounds that contain nitrogen but lack carbon cannot form $NaCN$.
Hydrazine hydrochloride $(NH_2NH_2 \cdot HCl)$ contains nitrogen but no carbon,therefore it fails the test.
Compounds that contain nitrogen but lack carbon cannot form $NaCN$.
Hydrazine hydrochloride $(NH_2NH_2 \cdot HCl)$ contains nitrogen but no carbon,therefore it fails the test.
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20
ChemistryMediumMCQAIPMT · 1994
In which of the following,addition of $HBr$ does not take place against Markownikoff's rule,or for which is the Anti-Markownikoff addition of $HBr$ not observed?
A
$Propene$
B
$But-1-ene$
C
$But-2-ene$
D
$Pent-2-ene$
Solution
(C) The Anti-Markownikoff addition of $HBr$ (also known as the peroxide effect or Kharasch effect) is only observed in unsymmetrical alkenes.
$But-2-ene$ $(CH_3-CH=CH-CH_3)$ is a symmetrical alkene.
In a symmetrical alkene,the addition of $HBr$ results in the same product regardless of whether it follows Markownikoff's rule or the Anti-Markownikoff mechanism.
Therefore,the Anti-Markownikoff addition is not observed for $But-2-ene$.
$But-2-ene$ $(CH_3-CH=CH-CH_3)$ is a symmetrical alkene.
In a symmetrical alkene,the addition of $HBr$ results in the same product regardless of whether it follows Markownikoff's rule or the Anti-Markownikoff mechanism.
Therefore,the Anti-Markownikoff addition is not observed for $But-2-ene$.
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21
ChemistryMediumMCQAIPMT · 1994
The attacking or reactive electrophilic species in the nitration of benzene with concentrated $HNO_3$ and $H_2SO_4$ is:
A
$NO_2^-$
B
$NO_2^+$
C
$NO_3^-$
D
$NO_2$
Solution
(B) In the nitration of benzene,the nitrating mixture (concentrated $HNO_3$ and $H_2SO_4$) reacts to produce the nitronium ion $(NO_2^+)$,which acts as the electrophile.
The reaction is: $HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + 2HSO_4^- + H_3O^+$
The reaction is: $HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + 2HSO_4^- + H_3O^+$
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22
ChemistryMCQAIPMT · 1994
Diazo-coupling is useful to prepare some
A
Dyes
B
Proteins
C
Pesticides
D
Vitamins
Solution
(A) The diazo-coupling reaction involves the reaction of a diazonium salt with an aromatic compound (like an amine or phenol) to form an azo compound.
These azo compounds are characterized by the $-N=N-$ group and are widely used as synthetic dyes due to their intense colors.
For example,the reaction between benzenediazonium chloride and aniline yields $p$-aminoazobenzene,which is a yellow dye.
Therefore,diazo-coupling is useful to prepare some dyes.
Correct option is $(A)$.
These azo compounds are characterized by the $-N=N-$ group and are widely used as synthetic dyes due to their intense colors.
For example,the reaction between benzenediazonium chloride and aniline yields $p$-aminoazobenzene,which is a yellow dye.
Therefore,diazo-coupling is useful to prepare some dyes.
Correct option is $(A)$.
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23
ChemistryMCQAIPMT · 1994
Pseudocoelom develops from
A
Blastopore lip
B
Archenteron
C
Embryonic mesoderm
D
Blastocoel
Solution
(D) The pseudocoelom is a body cavity that is not lined by mesoderm. During embryonic development,it arises from the persistent blastocoel. It is located between the body wall (ectoderm) and the digestive tract (endoderm),where the mesoderm is present only as scattered pouches rather than a continuous lining.
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24
ChemistryMCQAIPMT · 1994
In a stratified cambium,the fusiform initials are
A
Long and overlap each other at the ends
B
Short and overlap each other at the ends
C
Short and arranged in horizontal tiers
D
Short or long and overlap each other at the ends
Solution
(C) In a stratified cambium,the fusiform initials are relatively short and are arranged in horizontal tiers.
This arrangement is characteristic of certain plants where the initials are aligned in a regular,tiered pattern rather than overlapping at the ends as seen in non-stratified (storied) cambium.
This arrangement is characteristic of certain plants where the initials are aligned in a regular,tiered pattern rather than overlapping at the ends as seen in non-stratified (storied) cambium.
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25
ChemistryMCQAIPMT · 1994
Which one among the following is likely to have the highest levels of $DDT$ depositions in its body?
A
Eel
B
Crab
C
Phytoplankton
D
Sea gull
Solution
(D) The phenomenon of biomagnification refers to the increase in concentration of non-biodegradable substances (like $DDT$) at successive trophic levels.
Since $DDT$ is not metabolized or excreted, it accumulates in the fatty tissues of organisms.
In an aquatic food chain, the sequence is: $Phytoplankton \rightarrow Zooplankton \rightarrow Small \ fish \rightarrow Large \ fish \rightarrow Fish-eating \ birds$.
As the sea gull is at the top of this food chain, it consumes the highest quantity of $DDT$ accumulated from all the lower trophic levels, leading to the highest concentration of $DDT$ in its body.
Since $DDT$ is not metabolized or excreted, it accumulates in the fatty tissues of organisms.
In an aquatic food chain, the sequence is: $Phytoplankton \rightarrow Zooplankton \rightarrow Small \ fish \rightarrow Large \ fish \rightarrow Fish-eating \ birds$.
As the sea gull is at the top of this food chain, it consumes the highest quantity of $DDT$ accumulated from all the lower trophic levels, leading to the highest concentration of $DDT$ in its body.
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26
ChemistryMCQAIPMT · 1994
Which of the following instruments is used to measure the rate of transpiration in plants?
A
Porometer/Hygrometer
B
Potometer
C
Auxanometer
D
Tensiometer/Barometer
Solution
(B) $Potometer$ is an instrument used to measure the rate of transpiration in plants. It works on the principle that the amount of water absorbed by a plant is approximately equal to the amount of water lost through transpiration.
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27
ChemistryMCQAIPMT · 1994
Which of the following is a part of the pectoral girdle?
A
Coxal bone
B
Glenoid cavity
C
Acetabulum
D
Sternum
Solution
(B) The pectoral girdle (shoulder girdle) consists of two bones: the clavicle (collar bone) and the scapula (shoulder blade). The scapula contains a depression called the glenoid cavity,which articulates with the head of the humerus to form the shoulder joint. Therefore,the glenoid cavity is a part of the pectoral girdle. The coxal bone and acetabulum are parts of the pelvic girdle,while the sternum is part of the axial skeleton.
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28
ChemistryMCQAIPMT · 1994
Which of the following statements is correct regarding the cleavage of a human zygote?
A
It is meroblastic.
B
It starts when the ovum reaches the uterus.
C
It starts in the Fallopian tube.
D
It is identical to normal mitosis.
Solution
(C) मानव युग्मनज में विदलन (cleavage) होलोब्लास्टिक (holoblastic) प्रकार का होता है,जिसका अर्थ है कि यह पूरे युग्मनज में होता है। यह प्रक्रिया फैलोपियन ट्यूब में तब शुरू होती है जब युग्मनज गर्भाशय की ओर बढ़ रहा होता है। विदलन सामान्य समसूत्री विभाजन से भिन्न है क्योंकि इसमें कोशिकाओं की संख्या तो बढ़ती है लेकिन कोशिकाओं का आकार घटता जाता है और अंतरावस्था (interphase) बहुत कम समय की होती है।
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29
ChemistryMCQAIPMT · 1994
Which of the following has the largest population in a food chain?
A
Producers
B
Primary consumers
C
Secondary consumers
D
Decomposers
Solution
(A) In a typical food chain,the energy flow follows the $10\%$ law,where energy decreases at each trophic level.
Producers (autotrophs) form the base of the food chain and possess the highest biomass and population size to support the higher trophic levels.
As we move from producers to primary consumers,secondary consumers,and tertiary consumers,the number of individuals and the total biomass generally decrease.
Therefore,producers have the largest population in a food chain.
Producers (autotrophs) form the base of the food chain and possess the highest biomass and population size to support the higher trophic levels.
As we move from producers to primary consumers,secondary consumers,and tertiary consumers,the number of individuals and the total biomass generally decrease.
Therefore,producers have the largest population in a food chain.
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30
ChemistryMCQAIPMT · 1994
Evolutionary classification is called ........
A
Artificial system
B
Natural system
C
Phylogenetic system
D
None of the above
Solution
(C) Phylogenetic classification systems are based on evolutionary relationships between the various organisms. These systems assume that organisms belonging to the same taxa have a common ancestor. Therefore,evolutionary classification is known as the $Phylogenetic$ $system$.
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31
ChemistryMCQAIPMT · 1994
The total number of valence electrons in $4.2 \, g$ of $N_3^-$ ion is ......... $N_A$. ($N_A$ = Avogadro number)
A
$1.6$
B
$3.2$
C
$2.1$
D
$4.2$
Solution
(A) The molar mass of $N_3^-$ ion is $3 \times 14 = 42 \, g/mol$.
Number of moles of $N_3^-$ = $\frac{4.2 \, g}{42 \, g/mol} = 0.1 \, mol$.
Each $N$ atom has $5$ valence electrons. In $N_3^-$,total valence electrons = $(3 \times 5) + 1 = 16$.
Total valence electrons in $0.1 \, mol$ of $N_3^-$ = $0.1 \times 16 \times N_A = 1.6 \, N_A$.
Number of moles of $N_3^-$ = $\frac{4.2 \, g}{42 \, g/mol} = 0.1 \, mol$.
Each $N$ atom has $5$ valence electrons. In $N_3^-$,total valence electrons = $(3 \times 5) + 1 = 16$.
Total valence electrons in $0.1 \, mol$ of $N_3^-$ = $0.1 \times 16 \times N_A = 1.6 \, N_A$.
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32
ChemistryMCQAIPMT · 1994
Which of the following ligands is expected to be bidentate?
A
$Br^-$
B
$C_2O_4^{2-}$
C
$CH_3NH_2$
D
$CH_3C \equiv N$
Solution
(B) bidentate ligand is one that can coordinate to a central metal atom through two donor atoms.
$C_2O_4^{2-}$ (oxalate ion) has two oxygen donor atoms,making it a bidentate ligand.
$C_2O_4^{2-}$ (oxalate ion) has two oxygen donor atoms,making it a bidentate ligand.
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33
ChemistryMCQAIPMT · 1994
Which ion is essential for muscle contraction?
A
$Na^+$
B
$K^+$
C
$Ca^{++}$
D
$Cl^-$
Solution
(C) Muscle contraction is initiated by a signal sent by the central nervous system via a motor neuron.
When the action potential reaches the neuromuscular junction,it triggers the release of neurotransmitters,leading to an action potential in the sarcolemma.
This action potential spreads through the $T$-tubules and stimulates the sarcoplasmic reticulum to release calcium ions $(Ca^{++})$ into the sarcoplasm.
These $Ca^{++}$ ions bind to the troponin subunit on the actin filaments,which causes a conformational change that exposes the active sites for myosin binding.
Thus,$Ca^{++}$ is essential for the initiation of the cross-bridge cycle and muscle contraction.
When the action potential reaches the neuromuscular junction,it triggers the release of neurotransmitters,leading to an action potential in the sarcolemma.
This action potential spreads through the $T$-tubules and stimulates the sarcoplasmic reticulum to release calcium ions $(Ca^{++})$ into the sarcoplasm.
These $Ca^{++}$ ions bind to the troponin subunit on the actin filaments,which causes a conformational change that exposes the active sites for myosin binding.
Thus,$Ca^{++}$ is essential for the initiation of the cross-bridge cycle and muscle contraction.
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34
ChemistryMCQAIPMT · 1994
$A$ point source of light is placed $4 \, m$ below the surface of water of refractive index $5/3$. The minimum diameter of a disc that should be placed over the source on the surface of water to cut off all light coming out of the water is ...... $m$.
A
$2$
B
$4$
C
$3$
D
$6$
Solution
(D) The light from the point source will be cut off if the disc covers the area corresponding to the critical angle $\theta_c$ of the water-air interface.
For total internal reflection to occur,the radius $r$ of the disc is given by the formula $r = \frac{h}{\sqrt{\mu^2 - 1}}$,where $h = 4 \, m$ is the depth and $\mu = 5/3$ is the refractive index.
Substituting the values: $r = \frac{4}{\sqrt{(5/3)^2 - 1}} = \frac{4}{\sqrt{25/9 - 1}} = \frac{4}{\sqrt{16/9}} = \frac{4}{4/3} = 3 \, m$.
The diameter of the disc is $D = 2r = 2 \times 3 \, m = 6 \, m$.
For total internal reflection to occur,the radius $r$ of the disc is given by the formula $r = \frac{h}{\sqrt{\mu^2 - 1}}$,where $h = 4 \, m$ is the depth and $\mu = 5/3$ is the refractive index.
Substituting the values: $r = \frac{4}{\sqrt{(5/3)^2 - 1}} = \frac{4}{\sqrt{25/9 - 1}} = \frac{4}{\sqrt{16/9}} = \frac{4}{4/3} = 3 \, m$.
The diameter of the disc is $D = 2r = 2 \times 3 \, m = 6 \, m$.
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35
ChemistryMCQAIPMT · 1994
$A$ wire $50 \ cm$ long and $1 \ mm^2$ in cross-section carries a current of $4 \ A$ when connected to a $2 \ V$ battery. The resistivity of the wire is:
A
$4 \times 10^{-6} \ \Omega \cdot m$
B
$1 \times 10^{-6} \ \Omega \cdot m$
C
$2 \times 10^{-7} \ \Omega \cdot m$
D
$5 \times 10^{-7} \ \Omega \cdot m$
Solution
(B) Given: Length $\ell = 50 \ cm = 0.5 \ m$,Area $A = 1 \ mm^2 = 10^{-6} \ m^2$,Voltage $V = 2 \ V$,Current $I = 4 \ A$.
First,calculate the resistance $R$ using Ohm's Law: $R = \frac{V}{I} = \frac{2}{4} = 0.5 \ \Omega$.
The formula for resistivity $\rho$ is given by $R = \rho \frac{\ell}{A}$,which implies $\rho = \frac{R \cdot A}{\ell}$.
Substituting the values: $\rho = \frac{0.5 \times 10^{-6}}{0.5} = 1 \times 10^{-6} \ \Omega \cdot m$.
First,calculate the resistance $R$ using Ohm's Law: $R = \frac{V}{I} = \frac{2}{4} = 0.5 \ \Omega$.
The formula for resistivity $\rho$ is given by $R = \rho \frac{\ell}{A}$,which implies $\rho = \frac{R \cdot A}{\ell}$.
Substituting the values: $\rho = \frac{0.5 \times 10^{-6}}{0.5} = 1 \times 10^{-6} \ \Omega \cdot m$.
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36
ChemistryMediumMCQAIPMT · 1994
Which of the following statements concerning lanthanide elements is false?
A
Lanthanides are separated from one another by ion exchange method.
B
Ionic radii of trivalent lanthanides steadily increase with an increase in the atomic number.
C
All lanthanides are highly dense metals.
D
The most characteristic oxidation state of lanthanide elements is $+3$.
Solution
(B) The correct answer is $(B)$. The ionic radii of trivalent lanthanides do not increase with an increase in the atomic number; instead,they show a steady decrease,a phenomenon known as lanthanoid contraction.
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37
ChemistryEasyMCQAIPMT · 1994
Which one of the following statements is not correct?
A
Zinc dissolves in sodium hydroxide solution.
B
Carbon monoxide reduces iron $(III)$ oxide to iron.
C
Mercury $(II)$ iodide dissolves in excess of potassium iodide solution.
D
Tin $(IV)$ chloride is made by dissolving tin in concentrated hydrochloric acid.
Solution
(D) The correct statement analysis is as follows:
$1$. Zinc is amphoteric and reacts with $NaOH$ to form sodium zincate: $Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$. This statement is correct.
$2$. Carbon monoxide acts as a reducing agent in the blast furnace: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$. This statement is correct.
$3$. Mercury $(II)$ iodide dissolves in excess $KI$ to form a soluble complex: $HgI_2 + 2KI \rightarrow K_2[HgI_4]$. This statement is correct.
$4$. Tin $(IV)$ chloride $(SnCl_4)$ is prepared by the action of dry chlorine gas on tin metal $(Sn + 2Cl_2 \rightarrow SnCl_4)$. Dissolving tin in concentrated $HCl$ produces tin $(II)$ chloride $(SnCl_2)$,not tin $(IV)$ chloride. Therefore,this statement is incorrect.
$1$. Zinc is amphoteric and reacts with $NaOH$ to form sodium zincate: $Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$. This statement is correct.
$2$. Carbon monoxide acts as a reducing agent in the blast furnace: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$. This statement is correct.
$3$. Mercury $(II)$ iodide dissolves in excess $KI$ to form a soluble complex: $HgI_2 + 2KI \rightarrow K_2[HgI_4]$. This statement is correct.
$4$. Tin $(IV)$ chloride $(SnCl_4)$ is prepared by the action of dry chlorine gas on tin metal $(Sn + 2Cl_2 \rightarrow SnCl_4)$. Dissolving tin in concentrated $HCl$ produces tin $(II)$ chloride $(SnCl_2)$,not tin $(IV)$ chloride. Therefore,this statement is incorrect.
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38
ChemistryEasyMCQAIPMT · 1994
Which of the following elements does not form stable diatomic molecules?
A
Iodine
B
Phosphorus
C
Nitrogen
D
Oxygen
Solution
(B) Elements like $I_2$,$N_2$,and $O_2$ exist as stable diatomic molecules.
Phosphorus,due to its smaller size and high repulsion between lone pairs,prefers to form a tetrahedral $P_4$ molecule rather than a diatomic $P_2$ molecule.
Phosphorus,due to its smaller size and high repulsion between lone pairs,prefers to form a tetrahedral $P_4$ molecule rather than a diatomic $P_2$ molecule.
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39
ChemistryMediumMCQAIPMT · 1994
Which of the following oxides of nitrogen is paramagnetic?
A
$N_2O_3$
B
$N_2O$
C
$NO_2$
D
$N_2O_5$
Solution
(C) The electronic configuration of $NO_2$ involves a total of $17$ valence electrons.
Since the total number of valence electrons is odd,there is an unpaired electron present on the nitrogen atom.
Due to the presence of this unpaired electron,$NO_2$ is paramagnetic in nature.
Other oxides like $N_2O_3$,$N_2O$,and $N_2O_5$ have an even number of electrons and are diamagnetic.
Since the total number of valence electrons is odd,there is an unpaired electron present on the nitrogen atom.
Due to the presence of this unpaired electron,$NO_2$ is paramagnetic in nature.
Other oxides like $N_2O_3$,$N_2O$,and $N_2O_5$ have an even number of electrons and are diamagnetic.
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40
ChemistryMediumMCQAIPMT · 1994
How many grams of $CH_3OH$ should be added to water to prepare $150 \ mL$ solution of $2 \ M \ CH_3OH$?
A
$9.6$
B
$2.4$
C
$9.6 \times 10^3$
D
$2.4 \times 10^3$
Solution
(A) The molarity $(M)$ is defined as the number of moles of solute per liter of solution.
$M = \frac{\text{moles of solute}}{\text{volume of solution in } L}$.
Given: $M = 2 \ M$,Volume $(V)$ = $150 \ mL = 0.150 \ L$.
Moles of $CH_3OH = M \times V = 2 \ mol/L \times 0.150 \ L = 0.3 \ mol$.
Molar mass of $CH_3OH = (12 + 4 \times 1 + 16) = 32 \ g/mol$.
Mass of $CH_3OH = \text{moles} \times \text{molar mass} = 0.3 \ mol \times 32 \ g/mol = 9.6 \ g$.
$M = \frac{\text{moles of solute}}{\text{volume of solution in } L}$.
Given: $M = 2 \ M$,Volume $(V)$ = $150 \ mL = 0.150 \ L$.
Moles of $CH_3OH = M \times V = 2 \ mol/L \times 0.150 \ L = 0.3 \ mol$.
Molar mass of $CH_3OH = (12 + 4 \times 1 + 16) = 32 \ g/mol$.
Mass of $CH_3OH = \text{moles} \times \text{molar mass} = 0.3 \ mol \times 32 \ g/mol = 9.6 \ g$.
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41
ChemistryMediumMCQAIPMT · 1994
At $25 \ ^oC$,the highest osmotic pressure is exhibited by $0.1 \ M$ solution of
A
$CaCl_2$
B
$KCl$
C
Glucose
D
Urea
Solution
(A) The osmotic pressure $(\pi)$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
Since $C$,$R$,and $T$ are constant for all given solutions,the osmotic pressure depends directly on the van't Hoff factor $(i)$.
For $CaCl_2$,$i = 3$ $(Ca^{2+} + 2Cl^-)$.
For $KCl$,$i = 2$ $(K^+ + Cl^-)$.
For Glucose and Urea,$i = 1$ because they are non-electrolytes.
Since $CaCl_2$ has the highest value of $i$,it exhibits the highest osmotic pressure.
Since $C$,$R$,and $T$ are constant for all given solutions,the osmotic pressure depends directly on the van't Hoff factor $(i)$.
For $CaCl_2$,$i = 3$ $(Ca^{2+} + 2Cl^-)$.
For $KCl$,$i = 2$ $(K^+ + Cl^-)$.
For Glucose and Urea,$i = 1$ because they are non-electrolytes.
Since $CaCl_2$ has the highest value of $i$,it exhibits the highest osmotic pressure.
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42
ChemistryEasyMCQAIPMT · 1994
Which of the following salts has the same value of Van't Hoff factor $i$ as that of $K_4[Fe(CN)_6]$?
A
$Al_2(SO_4)_3$
B
$NaCl$
C
$Na_2SO_4$
D
$Al(NO_3)_3$
Solution
(A) The Van't Hoff factor $i$ represents the number of particles a solute dissociates into in a solution.
For $K_4[Fe(CN)_6]$,the dissociation is: $K_4[Fe(CN)_6] \rightarrow 4K^{+} + [Fe(CN)_6]^{4-}$.
Total number of particles $i = 4 + 1 = 5$.
Now,checking the options:
$A$. $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 2 + 3 = 5$.
$B$. $NaCl \rightarrow Na^{+} + Cl^{-}$,so $i = 1 + 1 = 2$.
$C$. $Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}$,so $i = 2 + 1 = 3$.
$D$. $Al(NO_3)_3 \rightarrow Al^{3+} + 3NO_3^{-}$,so $i = 1 + 3 = 4$.
Therefore,$Al_2(SO_4)_3$ has the same Van't Hoff factor as $K_4[Fe(CN)_6]$.
For $K_4[Fe(CN)_6]$,the dissociation is: $K_4[Fe(CN)_6] \rightarrow 4K^{+} + [Fe(CN)_6]^{4-}$.
Total number of particles $i = 4 + 1 = 5$.
Now,checking the options:
$A$. $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 2 + 3 = 5$.
$B$. $NaCl \rightarrow Na^{+} + Cl^{-}$,so $i = 1 + 1 = 2$.
$C$. $Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}$,so $i = 2 + 1 = 3$.
$D$. $Al(NO_3)_3 \rightarrow Al^{3+} + 3NO_3^{-}$,so $i = 1 + 3 = 4$.
Therefore,$Al_2(SO_4)_3$ has the same Van't Hoff factor as $K_4[Fe(CN)_6]$.
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43
ChemistryMediumMCQAIPMT · 1994
The number of atoms in $100 \ g$ of an $fcc$ crystal with density $d = 10 \ g/cm^3$ and cell edge equal to $100 \ pm$ is equal to
A
$4 \times 10^{25}$
B
$3 \times 10^{25}$
C
$2 \times 10^{25}$
D
$1 \times 10^{25}$
Solution
(A) For an $fcc$ crystal, the number of atoms per unit cell $(z)$ is $4$.
The volume of one unit cell is $V = a^3 = (100 \ pm)^3 = (100 \times 10^{-10} \ cm)^3 = 10^{-24} \ cm^3$.
The mass of one unit cell is $m = \text{density} \times \text{volume} = 10 \ g/cm^3 \times 10^{-24} \ cm^3 = 10^{-23} \ g$.
The number of unit cells in $100 \ g$ is $N_{cells} = \frac{100 \ g}{10^{-23} \ g/unit cell} = 10^{25} \ unit cells$.
Since each $fcc$ unit cell contains $4$ atoms, the total number of atoms is $4 \times 10^{25}$.
The volume of one unit cell is $V = a^3 = (100 \ pm)^3 = (100 \times 10^{-10} \ cm)^3 = 10^{-24} \ cm^3$.
The mass of one unit cell is $m = \text{density} \times \text{volume} = 10 \ g/cm^3 \times 10^{-24} \ cm^3 = 10^{-23} \ g$.
The number of unit cells in $100 \ g$ is $N_{cells} = \frac{100 \ g}{10^{-23} \ g/unit cell} = 10^{25} \ unit cells$.
Since each $fcc$ unit cell contains $4$ atoms, the total number of atoms is $4 \times 10^{25}$.
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44
ChemistryMediumMCQAIPMT · 1994
Ionic solids,with Schottky defects,contain in their structure
A
Equal number of cation and anion vacancies
B
Anion vacancies and interstitial anions
C
Cation vacancies only
D
Cation vacancies and interstitial cations
Solution
(A) The correct answer is $A$.
Schottky defect is a type of point defect in ionic crystals where an equal number of cations and anions are missing from their lattice sites,maintaining the electrical neutrality of the crystal.
Schottky defect is a type of point defect in ionic crystals where an equal number of cations and anions are missing from their lattice sites,maintaining the electrical neutrality of the crystal.
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45
ChemistryMediumMCQAIPMT · 1994
The data for the reaction $A + B \to C$ is given below. The rate law corresponding to the above data is:
| $Exp.$ | $[A]_0$ | $[B]_0$ | Initial rate |
|---|---|---|---|
| $(1)$ | $0.012$ | $0.035$ | $0.10$ |
| $(2)$ | $0.024$ | $0.070$ | $0.80$ |
| $(3)$ | $0.024$ | $0.035$ | $0.10$ |
| $(4)$ | $0.012$ | $0.070$ | $0.80$ |
A
Rate $= k[B]^3$
B
Rate $= k[B]^4$
C
Rate $= k[A][B]^3$
D
Rate $= k[A]^2[B]^2$
Solution
(A) Let the rate law be $Rate = k[A]^x[B]^y$.
From experiment $(1)$ and $(3)$:
$[A]$ changes from $0.012$ to $0.024$ (doubled),while $[B]$ is constant $(0.035)$. The rate remains $0.10$. Thus,$x = 0$.
From experiment $(1)$ and $(4)$:
$[A]$ is constant $(0.012)$,while $[B]$ changes from $0.035$ to $0.070$ (doubled). The rate changes from $0.10$ to $0.80$ (increased by $8$ times).
$2^y = 8 \implies y = 3$.
Therefore,the rate law is $Rate = k[B]^3$.
From experiment $(1)$ and $(3)$:
$[A]$ changes from $0.012$ to $0.024$ (doubled),while $[B]$ is constant $(0.035)$. The rate remains $0.10$. Thus,$x = 0$.
From experiment $(1)$ and $(4)$:
$[A]$ is constant $(0.012)$,while $[B]$ changes from $0.035$ to $0.070$ (doubled). The rate changes from $0.10$ to $0.80$ (increased by $8$ times).
$2^y = 8 \implies y = 3$.
Therefore,the rate law is $Rate = k[B]^3$.
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46
ChemistryMediumMCQAIPMT · 1994
Standard reduction potentials at $25^\circ C$ of $Li^{+}|Li$,$Ba^{2+}|Ba$,$Na^{+}|Na$,and $Mg^{2+}|Mg$ are $-3.05 \ V$,$-2.90 \ V$,$-2.71 \ V$,and $-2.37 \ V$ respectively. Which one of the following is the strongest oxidising agent?
A
$Na^{+}$
B
$Li^{+}$
C
$Ba^{2+}$
D
$Mg^{2+}$
Solution
(D) The strength of an oxidizing agent is directly proportional to its standard reduction potential $(E^\circ_{red})$.
Higher reduction potential indicates a greater tendency to gain electrons,making the species a stronger oxidizing agent.
Comparing the given values: $E^\circ(Li^{+}|Li) = -3.05 \ V$,$E^\circ(Ba^{2+}|Ba) = -2.90 \ V$,$E^\circ(Na^{+}|Na) = -2.71 \ V$,and $E^\circ(Mg^{2+}|Mg) = -2.37 \ V$.
Since $-2.37 \ V$ is the highest value among the given potentials,$Mg^{2+}$ is the strongest oxidizing agent.
Higher reduction potential indicates a greater tendency to gain electrons,making the species a stronger oxidizing agent.
Comparing the given values: $E^\circ(Li^{+}|Li) = -3.05 \ V$,$E^\circ(Ba^{2+}|Ba) = -2.90 \ V$,$E^\circ(Na^{+}|Na) = -2.71 \ V$,and $E^\circ(Mg^{2+}|Mg) = -2.37 \ V$.
Since $-2.37 \ V$ is the highest value among the given potentials,$Mg^{2+}$ is the strongest oxidizing agent.
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47
ChemistryMediumMCQAIPMT · 1994
Which of the following displaces $Br_2$ from an aqueous solution containing bromide ions?
A
$Cl_2$
B
$Cl^-$
C
$I_2$
D
$I_3^-$
Solution
(A) The displacement of a halide ion by a halogen depends on their standard reduction potentials.
According to the electrochemical series,the oxidizing power of halogens decreases in the order: $F_2 > Cl_2 > Br_2 > I_2$.
Since $Cl_2$ has a higher reduction potential than $Br_2$,it can oxidize bromide ions $(Br^-)$ to bromine $(Br_2)$ according to the reaction: $Cl_2(g) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(l)$.
Therefore,$Cl_2$ is the correct answer.
According to the electrochemical series,the oxidizing power of halogens decreases in the order: $F_2 > Cl_2 > Br_2 > I_2$.
Since $Cl_2$ has a higher reduction potential than $Br_2$,it can oxidize bromide ions $(Br^-)$ to bromine $(Br_2)$ according to the reaction: $Cl_2(g) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(l)$.
Therefore,$Cl_2$ is the correct answer.
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48
ChemistryEasyMCQAIPMT · 1994
For the adsorption of a gas on a solid,the plot of $\log(x/m)$ versus $\log P$ is linear with slope equal to
A
$k$
B
$\log k$
C
$n$
D
$1/n$
Solution
(D) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = kp^{1/n}$.
Taking the logarithm on both sides,we get: $\log(x/m) = \log k + \frac{1}{n} \log p$.
Comparing this with the linear equation $y = mx + c$,where $y = \log(x/m)$,$x = \log p$,$m = \text{slope} = 1/n$,and $c = \text{intercept} = \log k$.
Therefore,the slope of the plot of $\log(x/m)$ versus $\log p$ is $1/n$.
Taking the logarithm on both sides,we get: $\log(x/m) = \log k + \frac{1}{n} \log p$.
Comparing this with the linear equation $y = mx + c$,where $y = \log(x/m)$,$x = \log p$,$m = \text{slope} = 1/n$,and $c = \text{intercept} = \log k$.
Therefore,the slope of the plot of $\log(x/m)$ versus $\log p$ is $1/n$.
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49
ChemistryEasyMCQAIPMT · 1994
The method used for obtaining highly pure silicon,which is used as a semiconductor material,is:
A
Oxidation
B
Electrochemical
C
Crystallization
D
Zone refining
Solution
(D) Zone refining is the method used for obtaining metals and semiconductors in a state of high purity. In this process,a circular mobile heater is fixed at one end of a rod of the impure metal. As the heater moves,the pure metal crystallizes out of the melt and the impurities pass into the adjacent molten zone. This process is repeated several times to obtain high-purity silicon.
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50
ChemistryEasyMCQAIPMT · 1994
Highest $(+7)$ oxidation state is shown by
A
$Co$
B
$Cr$
C
$V$
D
$Mn$
Solution
(D) $Mn$ (Manganese) belongs to group $7$ of the periodic table. Its electronic configuration is $[Ar] 3d^5 4s^2$. It can lose all $7$ valence electrons to exhibit a maximum oxidation state of $+7$,as seen in compounds like $KMnO_4$.
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51
ChemistryEasyMCQAIPMT · 1994
To protect iron against corrosion,the most durable metal plating on it is:
A
Nickel plating
B
Tin plating
C
Copper plating
D
Zinc plating
Solution
(D) Galvanisation is the process of applying a protective $Zn$ coating to iron or steel to prevent rusting.
Since $Zn$ is more reactive than $Fe$,it acts as a sacrificial anode and protects the iron even if the coating is scratched.
Therefore,$Zn$ plating is the most durable method for protecting iron against corrosion.
Since $Zn$ is more reactive than $Fe$,it acts as a sacrificial anode and protects the iron even if the coating is scratched.
Therefore,$Zn$ plating is the most durable method for protecting iron against corrosion.
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52
ChemistryDifficultMCQAIPMT · 1994
Which of the following ligands is expected to be bidentate?
A
$Br^{-}$
B
$C_2O_4^{2-}$
C
$CH_3NH_2$
D
$CH_3C \equiv N$
Solution
(B)
Ligands that have two donor atoms to coordinate with the central metal ion are called bidentate ligands.
$C_2O_4^{2-}$ (oxalate ion) has two oxygen donor atoms and is a bidentate ligand.
Ligands that have two donor atoms to coordinate with the central metal ion are called bidentate ligands.
$C_2O_4^{2-}$ (oxalate ion) has two oxygen donor atoms and is a bidentate ligand.
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53
ChemistryMediumMCQAIPMT · 1994
Grignard reagent is prepared by the reaction between
A
Zinc and alkyl halide
B
Magnesium and alkyl halide
C
Magnesium and alkane
D
Magnesium and aromatic hydrocarbon
Solution
(B) is the correct answer.
Grignard reagent $(RMgX)$ is prepared by the reaction of an alkyl halide $(RX)$ with magnesium metal in the presence of dry ether.
The reaction is: $RX + Mg \xrightarrow{\text{Dry ether}} RMgX$.
Grignard reagent $(RMgX)$ is prepared by the reaction of an alkyl halide $(RX)$ with magnesium metal in the presence of dry ether.
The reaction is: $RX + Mg \xrightarrow{\text{Dry ether}} RMgX$.
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54
ChemistryDifficultMCQAIPMT · 1994
Reaction of $t-$butyl bromide with sodium methoxide produces
A
Isobutane
B
Isobutylene
C
Sodium $t-$butoxide
D
$t-$butyl methyl ether
Solution
(B) $(CH_3)_3CBr + CH_3ONa \xrightarrow{\text{Elimination}} CH_2=C(CH_3)_2 + CH_3OH + NaBr$
The product is Isobutylene.
$CH_3ONa \to CH_3O^{-} + Na^{+}$
Methoxide ion $(CH_3O^{-})$ is a strong base,therefore it abstracts a proton from the $3^{\circ}$ alkyl halide and favours an elimination reaction.
The product is Isobutylene.
$CH_3ONa \to CH_3O^{-} + Na^{+}$
Methoxide ion $(CH_3O^{-})$ is a strong base,therefore it abstracts a proton from the $3^{\circ}$ alkyl halide and favours an elimination reaction.
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55
ChemistryAdvancedMCQAIPMT · 1994
$A$ compound $A$ has a molecular formula $C_2Cl_3OH$. It reduces Fehling's solution and on oxidation gives a monocarboxylic acid $(B)$. $A$ is obtained by the action of chlorine on ethyl alcohol. $A$ is:
A
Chloral
B
$CHCl_3$
C
$CH_3Cl$
D
Chloroacetic acid
Solution
(A) The molecular formula $C_2Cl_3OH$ can be written as $CCl_3CHO$ (Chloral).
Since $A$ reduces Fehling's solution,it must contain an aldehyde $(-CHO)$ group.
Chloral $(CCl_3CHO)$ is formed by the chlorination of ethyl alcohol $(C_2H_5OH)$:
$C_2H_5OH + 4Cl_2 \to CCl_3CHO + 5HCl$.
Upon oxidation,Chloral gives trichloroacetic acid $(CCl_3COOH)$,which is a monocarboxylic acid $(B)$.
Therefore,$A$ is Chloral.
Since $A$ reduces Fehling's solution,it must contain an aldehyde $(-CHO)$ group.
Chloral $(CCl_3CHO)$ is formed by the chlorination of ethyl alcohol $(C_2H_5OH)$:
$C_2H_5OH + 4Cl_2 \to CCl_3CHO + 5HCl$.
Upon oxidation,Chloral gives trichloroacetic acid $(CCl_3COOH)$,which is a monocarboxylic acid $(B)$.
Therefore,$A$ is Chloral.
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56
ChemistryMediumMCQAIPMT · 1994
Diazo-coupling is useful to prepare some
A
Pesticides
B
Proteins
C
Dyes
D
Vitamins
Solution
(C) Diazo-coupling reactions involve the reaction of a diazonium salt with an aromatic compound (like phenols or amines) to form azo compounds.
These azo compounds are characterized by the $-N=N-$ linkage and are extensively used as synthetic dyes due to their intense colors.
For example,the reaction of benzene diazonium chloride with aniline produces $p$-aminoazobenzene,which is a yellow dye.
These azo compounds are characterized by the $-N=N-$ linkage and are extensively used as synthetic dyes due to their intense colors.
For example,the reaction of benzene diazonium chloride with aniline produces $p$-aminoazobenzene,which is a yellow dye.
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57
ChemistryDifficultMCQAIPMT · 1994
On heating glycerol with conc. $H_2SO_4$,a compound is obtained which has a bad odour. The compound is:
A
Glycerol sulphate
B
Acrolein
C
Formic acid
D
Allyl alcohol
Solution
(B) . When glycerol is heated with a dehydrating agent like concentrated $H_2SO_4$ or $KHSO_4$,it undergoes dehydration to form acrolein (propenal),which has a characteristic pungent (bad) odour.
$CH_2OH-CHOH-CH_2OH \xrightarrow{\text{conc. } H_2SO_4, \Delta} CH_2=CH-CHO + 2H_2O$
$CH_2OH-CHOH-CH_2OH \xrightarrow{\text{conc. } H_2SO_4, \Delta} CH_2=CH-CHO + 2H_2O$
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58
ChemistryDifficultMCQAIPMT · 1994
The compound which does not react with sodium is
A
$C_2H_5OH$
B
$CH_3-O-CH_3$
C
$CH_3COOH$
D
$CH_3-CH(OH)-CH_3$
Solution
(B) Sodium $(Na)$ reacts with compounds containing acidic hydrogen atoms (like $-OH$ or $-COOH$ groups) to release hydrogen gas $(H_2)$.
$C_2H_5OH$ (ethanol),$CH_3COOH$ (acetic acid),and $CH_3-CH(OH)-CH_3$ (propan$-2-$ol) all contain an acidic hydrogen atom attached to an oxygen atom.
$CH_3-O-CH_3$ (dimethyl ether) is an ether and does not contain any replaceable hydrogen atom attached to an oxygen atom.
Therefore,it does not react with sodium metal.
$C_2H_5OH$ (ethanol),$CH_3COOH$ (acetic acid),and $CH_3-CH(OH)-CH_3$ (propan$-2-$ol) all contain an acidic hydrogen atom attached to an oxygen atom.
$CH_3-O-CH_3$ (dimethyl ether) is an ether and does not contain any replaceable hydrogen atom attached to an oxygen atom.
Therefore,it does not react with sodium metal.
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59
ChemistryMediumMCQAIPMT · 1994
The most suitable method for the separation of a $1:1$ mixture of $o-$ and $p-$ nitrophenols is:
A
Steam distillation
B
Sublimation
C
Crystallization
D
Chromatography
Solution
(A) $o-$ and $p-$ nitrophenols are separated by steam distillation.
$o-$ nitrophenol exhibits intramolecular hydrogen bonding,making it steam volatile.
$p-$ nitrophenol exhibits intermolecular hydrogen bonding,leading to higher molecular association and lower volatility,thus it is not steam volatile.
$o-$ nitrophenol exhibits intramolecular hydrogen bonding,making it steam volatile.
$p-$ nitrophenol exhibits intermolecular hydrogen bonding,leading to higher molecular association and lower volatility,thus it is not steam volatile.
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60
ChemistryMediumMCQAIPMT · 1994
Which of the following compounds will undergo self-aldol condensation in the presence of cold dilute alkali?
A
$C_6H_5CHO$
B
$CH_3CH_2CHO$
C
$CH_3CHO$
D
$HCHO$
Solution
(B) Aldol condensation occurs in aldehydes or ketones that possess at least one $\alpha$-hydrogen atom.
$C_6H_5CHO$ (benzaldehyde) and $HCHO$ (formaldehyde) do not have $\alpha$-hydrogen atoms.
$CH_3CH_2CHO$ (propanal) contains two $\alpha$-hydrogen atoms attached to the $\alpha$-carbon.
Therefore,$CH_3CH_2CHO$ undergoes self-aldol condensation in the presence of cold dilute alkali.
$C_6H_5CHO$ (benzaldehyde) and $HCHO$ (formaldehyde) do not have $\alpha$-hydrogen atoms.
$CH_3CH_2CHO$ (propanal) contains two $\alpha$-hydrogen atoms attached to the $\alpha$-carbon.
Therefore,$CH_3CH_2CHO$ undergoes self-aldol condensation in the presence of cold dilute alkali.
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61
ChemistryDifficultMCQAIPMT · 1994
An acyl halide is formed when $PCl_5$ reacts with an
A
Acid
B
Alcohol
C
Amide
D
Ester
Solution
(A) The reaction of a carboxylic acid with $PCl_5$ produces an acyl chloride (acid chloride).
The chemical equation is: $CH_3COOH + PCl_5 \to CH_3COCl + POCl_3 + HCl$
Therefore,the correct option is $A$.
The chemical equation is: $CH_3COOH + PCl_5 \to CH_3COCl + POCl_3 + HCl$
Therefore,the correct option is $A$.
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62
ChemistryDifficultMCQAIPMT · 1994
Which one of the following orders is wrong with respect to the property indicated?
A
Formic acid $ > $ acetic acid $ > $ propanoic acid (acid strength)
B
Fluoroacetic acid $ > $ chloroacetic acid $ > $ bromoacetic acid (acid strength)
C
Benzoic acid $ > $ phenol $ > $ cyclohexanol (acid strength)
D
Aniline $ > $ cyclohexylamine $ > $ benzamide (basic strength)
Solution
(D) The correct order of basic strength is: $\text{cyclohexylamine} > \text{aniline} > \text{benzamide}$.
In $\text{cyclohexylamine}$,the nitrogen lone pair is localized.
In $\text{aniline}$,the lone pair is involved in resonance with the benzene ring,reducing basicity.
In $\text{benzamide}$,the lone pair is involved in resonance with both the benzene ring and the carbonyl group,significantly reducing basicity.
Therefore,the order given in option $D$ is incorrect.
In $\text{cyclohexylamine}$,the nitrogen lone pair is localized.
In $\text{aniline}$,the lone pair is involved in resonance with the benzene ring,reducing basicity.
In $\text{benzamide}$,the lone pair is involved in resonance with both the benzene ring and the carbonyl group,significantly reducing basicity.
Therefore,the order given in option $D$ is incorrect.
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63
ChemistryMediumMCQAIPMT · 1994
Reaction of nitrous acid with aliphatic primary amine in the cold gives
A
$A$. $A$ diazonium salt
B
$B$. An alcohol
C
$C$. $A$ nitrite
D
$D$. $A$ dye
Solution
(B) When an aliphatic primary amine $(R-NH_2)$ reacts with nitrous acid $(HNO_2)$ at low temperatures (cold conditions),it forms an unstable aliphatic diazonium salt,which immediately decomposes to evolve nitrogen gas $(N_2)$ and forms an alcohol $(R-OH)$.
The reaction is as follows:
$R-NH_2 + HNO_2$ $\xrightarrow{\text{cold}} [R-N_2^+Cl^-]$ $\rightarrow R-OH + N_2 + H_2O$
The reaction is as follows:
$R-NH_2 + HNO_2$ $\xrightarrow{\text{cold}} [R-N_2^+Cl^-]$ $\rightarrow R-OH + N_2 + H_2O$
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64
ChemistryEasyMCQAIPMT · 1994
By the action of enzymes,the rate of biochemical reaction
A
Decreases
B
Increases
C
Does not change
D
Either $(a)$ or $(c)$
Solution
(B) Enzymes are highly efficient catalysts for biochemical reactions.
They speed up the reaction by providing an alternative pathway with lower activation energy.
They speed up the reaction by providing an alternative pathway with lower activation energy.
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65
ChemistryMediumMCQAIPMT · 1994
Which statement is false?
A
Some disinfectants can be used as antiseptics at low concentration.
B
Sulphadiazine is a synthetic antibacterial.
C
Ampicillin is a natural antibiotic.
D
Aspirin is analgesic and antipyretic both.
Solution
(C) is the false statement. Ampicillin is a synthetic modification of penicillin,not a natural antibiotic.
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66
ChemistryEasyMCQAIPMT · 1994
Which of the following compounds will give a positive test with Tollen's reagent?
A
Acetamide
B
Acetaldehyde
C
Acetic acid
D
Acetone
Solution
(B) Tollen's reagent is an ammoniacal silver nitrate solution. Its active oxidizing species is $Ag^+$. It oxidizes both aliphatic and aromatic aldehydes to their corresponding carboxylic acids,while $Ag^+$ is reduced to metallic silver $(Ag)$,forming a silver mirror.The reaction is: $R-CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag(s) + 4NH_3 + 2H_2O$.Among the given options,$Acetaldehyde$ $(CH_3CHO)$ is an aldehyde and therefore gives a positive Tollen's test.
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