ChemistryQ151–157 of 183 questions
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151
ChemistryEasyMCQTS EAMCET · 2004
An organic compound containing $C$ and $H$ has $92.3 \%$ of carbon. Its empirical formula is:
A
$CH$
B
$CH_3$
C
$CH_2$
D
$CH_4$
Solution
(A) The percentage of $C$ is $92.3 \%$,so the percentage of $H$ is $100 - 92.3 = 7.7 \%$.
Thus,the empirical formula is $CH$.
| Element | Atomic Ratio (Percentage / Atomic Mass) | Simple Molar Ratio |
| $C$ | $92.3 / 12 = 7.69$ | $7.69 / 7.69 = 1$ |
| $H$ | $7.7 / 1 = 7.70$ | $7.70 / 7.69 \approx 1$ |
Thus,the empirical formula is $CH$.
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152
ChemistryDifficultMCQTS EAMCET · 2004
At $27^{\circ} C$,$500 \ mL$ of helium diffuses in $30 \ minutes$. What is the time (in hours) taken for $1000 \ mL$ of $SO_2$ to diffuse under the same experimental conditions?
A
$240$
B
$3$
C
$2$
D
$4$
Solution
(D) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ and directly proportional to the volume $V$ diffused per unit time $t$: $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$.
Given for Helium $(He)$: $V_1 = 500 \ mL$,$t_1 = 30 \ min$,$M_1 = 4 \ g/mol$.
Given for Sulfur dioxide $(SO_2)$: $V_2 = 1000 \ mL$,$t_2 = t$,$M_2 = 64 \ g/mol$.
The ratio is given by: $\frac{r_1}{r_2} = \frac{V_1/t_1}{V_2/t_2} = \sqrt{\frac{M_2}{M_1}}$.
Substituting the values: $\frac{500/30}{1000/t} = \sqrt{\frac{64}{4}}$.
$\frac{500}{30} \times \frac{t}{1000} = \sqrt{16} = 4$.
$\frac{t}{60} = 4$.
$t = 240 \ minutes$.
Converting to hours: $t = \frac{240}{60} = 4 \ hours$.
Given for Helium $(He)$: $V_1 = 500 \ mL$,$t_1 = 30 \ min$,$M_1 = 4 \ g/mol$.
Given for Sulfur dioxide $(SO_2)$: $V_2 = 1000 \ mL$,$t_2 = t$,$M_2 = 64 \ g/mol$.
The ratio is given by: $\frac{r_1}{r_2} = \frac{V_1/t_1}{V_2/t_2} = \sqrt{\frac{M_2}{M_1}}$.
Substituting the values: $\frac{500/30}{1000/t} = \sqrt{\frac{64}{4}}$.
$\frac{500}{30} \times \frac{t}{1000} = \sqrt{16} = 4$.
$\frac{t}{60} = 4$.
$t = 240 \ minutes$.
Converting to hours: $t = \frac{240}{60} = 4 \ hours$.
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153
ChemistryEasyMCQTS EAMCET · 2004
Assertion $(A)$ At $300 \ K$,kinetic energy of $16 \ g$ of methane is equal to the kinetic energy of $32 \ g$ of oxygen.
Reason $(R)$ At constant temperature,kinetic energy of one mole of all gases is equal.
Reason $(R)$ At constant temperature,kinetic energy of one mole of all gases is equal.
A
Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$
B
Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A)$
C
$(A)$ is true but $(R)$ is not true
D
$(A)$ is not true but $(R)$ is true
Solution
(A) The kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by $KE = \frac{3}{2} nRT$.
For $16 \ g$ of methane ($CH_4$,molar mass = $16 \ g/mol$),$n = \frac{16}{16} = 1 \ mol$.
For $32 \ g$ of oxygen ($O_2$,molar mass = $32 \ g/mol$),$n = \frac{32}{32} = 1 \ mol$.
Since both samples contain $1 \ mol$ of gas at the same temperature $(300 \ K)$,their kinetic energies are equal. Thus,$(A)$ is true.
The kinetic energy of $1 \ mol$ of any ideal gas is $\frac{3}{2} RT$,which depends only on temperature. Thus,$(R)$ is true and correctly explains $(A)$.
For $16 \ g$ of methane ($CH_4$,molar mass = $16 \ g/mol$),$n = \frac{16}{16} = 1 \ mol$.
For $32 \ g$ of oxygen ($O_2$,molar mass = $32 \ g/mol$),$n = \frac{32}{32} = 1 \ mol$.
Since both samples contain $1 \ mol$ of gas at the same temperature $(300 \ K)$,their kinetic energies are equal. Thus,$(A)$ is true.
The kinetic energy of $1 \ mol$ of any ideal gas is $\frac{3}{2} RT$,which depends only on temperature. Thus,$(R)$ is true and correctly explains $(A)$.
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154
ChemistryEasyMCQTS EAMCET · 2004
Which of the following is correct?
A
$^1_1 H$ and $^3_2 He$ are isotopes
B
$^{14}_6 C$ and $^{14}_7 N$ are isotopes
C
$^{39}_{19} K$ and $^{40}_{20} Ca$ are isotones
D
$^{19}_9 F$ and $^{24}_{11} Na$ are isotopes
Solution
(C) Isotones are species that have an equal number of neutrons.
Number of neutrons = Mass number $(A)$ - Atomic number $(Z)$.
For $^{39}_{19} K$: Neutrons = $39 - 19 = 20$.
For $^{40}_{20} Ca$: Neutrons = $40 - 20 = 20$.
Since both have $20$ neutrons,they are isotones.
Number of neutrons = Mass number $(A)$ - Atomic number $(Z)$.
For $^{39}_{19} K$: Neutrons = $39 - 19 = 20$.
For $^{40}_{20} Ca$: Neutrons = $40 - 20 = 20$.
Since both have $20$ neutrons,they are isotones.
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155
ChemistryMediumMCQTS EAMCET · 2004
Which of the following elements has the least number of electrons in its $M$ shell?
A
$K$
B
$Mn$
C
$Ni$
D
$Sc$
Solution
(A) The electronic configurations of the given elements are as follows:
$K (Z=19): 1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^1$. The $M$ shell $(n=3)$ has $2+6 = 8$ electrons.
$Mn (Z=25): 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^5, 4s^2$. The $M$ shell $(n=3)$ has $2+6+5 = 13$ electrons.
$Ni (Z=28): 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^8, 4s^2$. The $M$ shell $(n=3)$ has $2+6+8 = 16$ electrons.
$Sc (Z=21): 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^1, 4s^2$. The $M$ shell $(n=3)$ has $2+6+1 = 9$ electrons.
Comparing the number of electrons in the $M$ shell: $K (8) < Sc (9) < Mn (13) < Ni (16)$.
Thus,$K$ has the least number of electrons in its $M$ shell.
$K (Z=19): 1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^1$. The $M$ shell $(n=3)$ has $2+6 = 8$ electrons.
$Mn (Z=25): 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^5, 4s^2$. The $M$ shell $(n=3)$ has $2+6+5 = 13$ electrons.
$Ni (Z=28): 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^8, 4s^2$. The $M$ shell $(n=3)$ has $2+6+8 = 16$ electrons.
$Sc (Z=21): 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^1, 4s^2$. The $M$ shell $(n=3)$ has $2+6+1 = 9$ electrons.
Comparing the number of electrons in the $M$ shell: $K (8) < Sc (9) < Mn (13) < Ni (16)$.
Thus,$K$ has the least number of electrons in its $M$ shell.
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156
ChemistryEasyMCQTS EAMCET · 2004
Which of the following is an endothermic reaction?
A
$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)} + 92 \ kJ$
B
$N_{2(g)} + O_{2(g)} + 180.8 \ kJ \longrightarrow 2 NO_{(g)}$
C
$H_{2(g)} + Cl_{2(g)} \longrightarrow 2 HCl_{(g)} + 184.6 \ kJ$
D
$C_{(graphite)} + 2 H_{2(g)} \longrightarrow CH_{4(g)} + 74.8 \ kJ$
Solution
(B) An endothermic reaction is a process in which heat energy is absorbed from the surroundings.
In the reaction $N_{2(g)} + O_{2(g)} + 180.8 \ kJ \longrightarrow 2 NO_{(g)}$,heat is added as a reactant,indicating that energy is absorbed.
Therefore,this is an endothermic reaction.
In the reaction $N_{2(g)} + O_{2(g)} + 180.8 \ kJ \longrightarrow 2 NO_{(g)}$,heat is added as a reactant,indicating that energy is absorbed.
Therefore,this is an endothermic reaction.
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157
ChemistryMediumMCQTS EAMCET · 2004
Average $C-H$ bond energy is $416 \ kJ \ mol^{-1}$. Which of the following is correct?
A
$CH_{4(g)} + 416 \ kJ \longrightarrow C_{(g)} + 4H_{(g)}$
B
$CH_{4(g)} \longrightarrow C_{(g)} + 4H_{(g)} + 416 \ kJ$
C
$CH_{4(g)} + 1664 \ kJ \longrightarrow C_{(g)} + 4H_{(g)}$
D
$CH_{4(g)} \longrightarrow C_{(g)} + 4H_{(g)} + 1664 \ kJ$
Solution
(C) The molecule $CH_4$ contains four $C-H$ bonds.
To break one mole of $CH_4$ into gaseous carbon and hydrogen atoms,all four $C-H$ bonds must be broken.
The total energy required is $4 \times 416 \ kJ \ mol^{-1} = 1664 \ kJ \ mol^{-1}$.
Therefore,the thermochemical equation is $CH_{4(g)} + 1664 \ kJ \longrightarrow C_{(g)} + 4H_{(g)}$.
To break one mole of $CH_4$ into gaseous carbon and hydrogen atoms,all four $C-H$ bonds must be broken.
The total energy required is $4 \times 416 \ kJ \ mol^{-1} = 1664 \ kJ \ mol^{-1}$.
Therefore,the thermochemical equation is $CH_{4(g)} + 1664 \ kJ \longrightarrow C_{(g)} + 4H_{(g)}$.
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