ChemistryQ1–30 of 30 questions
Page 1 of 1 · English
1
ChemistryMediumMCQIIT JEE · 1984
When $\alpha$-particles are sent through a thin metal foil,most of them go straight through the foil because (one or more are correct)
A
Alpha particles are much heavier than electrons
B
Alpha particles are positively charged
C
Most part of the atom is empty space
D
Alpha particles move with high velocity
Solution
(C) The correct answer is $(C)$.
In the Rutherford $\alpha$-particle scattering experiment,it was observed that most of the $\alpha$-particles passed through the gold foil without any deflection.
This indicates that the majority of the space within an atom is empty.
In the Rutherford $\alpha$-particle scattering experiment,it was observed that most of the $\alpha$-particles passed through the gold foil without any deflection.
This indicates that the majority of the space within an atom is empty.
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2
ChemistryMediumMCQIIT JEE · 1984
The correct set of four quantum numbers for the valence electron of rubidium $(Z = 37)$ is:
A
$5, 0, 0, +\frac{1}{2}$
B
$5, 1, 0, +\frac{1}{2}$
C
$5, 1, 1, +\frac{1}{2}$
D
$6, 0, 0, +\frac{1}{2}$
Solution
(A) The atomic number of rubidium $(Rb)$ is $37$.
The electronic configuration of $Rb$ is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 5s^1$.
The valence electron is in the $5s$ orbital.
For the $5s^1$ electron:
Principal quantum number $(n) = 5$.
Azimuthal quantum number $(l)$ for $s$-orbital $= 0$.
Magnetic quantum number $(m_l) = 0$.
Spin quantum number $(m_s) = +\frac{1}{2}$ (or $-\frac{1}{2}$).
Thus,the set of quantum numbers is $(5, 0, 0, +\frac{1}{2})$.
The electronic configuration of $Rb$ is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 5s^1$.
The valence electron is in the $5s$ orbital.
For the $5s^1$ electron:
Principal quantum number $(n) = 5$.
Azimuthal quantum number $(l)$ for $s$-orbital $= 0$.
Magnetic quantum number $(m_l) = 0$.
Spin quantum number $(m_s) = +\frac{1}{2}$ (or $-\frac{1}{2}$).
Thus,the set of quantum numbers is $(5, 0, 0, +\frac{1}{2})$.
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3
ChemistryMediumMCQIIT JEE · 1984
The increasing order (lowest first) for the values of $e/m$ (charge/mass) for:
A
$e, p, n, \alpha$
B
$n, p, e, \alpha$
C
$n, p, \alpha, e$
D
$n, \alpha, p, e$
Solution
(D) The $e/m$ ratio (specific charge) is calculated as follows:
$(I)$ For neutron $(n)$: Charge = $0$,Mass = $1$,so $e/m = 0/1 = 0$.
$(II)$ For $\alpha$-particle $(\alpha)$: Charge = $ 2$,Mass = $4$,so $e/m = 2/4 = 0.5$.
$(III)$ For proton $(p)$: Charge = $ 1$,Mass = $1$,so $e/m = 1/1 = 1$.
$(IV)$ For electron $(e)$: Charge = $-1$,Mass = $1/1837$,so $e/m = 1/(1/1837) = 1837$.
Comparing these values: $0 < 0.5 < 1 < 1837$.
Therefore,the increasing order is $n < \alpha < p < e$.
$(I)$ For neutron $(n)$: Charge = $0$,Mass = $1$,so $e/m = 0/1 = 0$.
$(II)$ For $\alpha$-particle $(\alpha)$: Charge = $ 2$,Mass = $4$,so $e/m = 2/4 = 0.5$.
$(III)$ For proton $(p)$: Charge = $ 1$,Mass = $1$,so $e/m = 1/1 = 1$.
$(IV)$ For electron $(e)$: Charge = $-1$,Mass = $1/1837$,so $e/m = 1/(1/1837) = 1837$.
Comparing these values: $0 < 0.5 < 1 < 1837$.
Therefore,the increasing order is $n < \alpha < p < e$.
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4
ChemistryEasyMCQIIT JEE · 1984
Oxygen molecule is paramagnetic because
A
Bonding electrons are more than antibonding electrons
B
Contains unpaired electrons
C
Bonding electrons are less than antibonding electrons
D
Bonding electrons are equal to antibonding electrons
Solution
(B) According to $MOT$,the electronic configuration of $O_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since it contains $2$ unpaired electrons in the $\pi^*$ antibonding molecular orbitals,it is paramagnetic.
Since it contains $2$ unpaired electrons in the $\pi^*$ antibonding molecular orbitals,it is paramagnetic.
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5
ChemistryEasyMCQIIT JEE · 1984
As a result of $sp$ hybridization,we get
A
Two mutually perpendicular orbitals
B
Two orbitals at $180^\circ$
C
Four orbitals in tetrahedral directions
D
Three orbitals in the same plane
Solution
(B) $sp$-hybridization involves the mixing of one $s$ and one $p$ orbital to form two equivalent $sp$ hybrid orbitals.
These two hybrid orbitals are oriented at an angle of $180^\circ$ to each other,resulting in a linear geometry.
These two hybrid orbitals are oriented at an angle of $180^\circ$ to each other,resulting in a linear geometry.
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6
ChemistryDifficultMCQIIT JEE · 1984
When an ideal gas undergoes unrestrained expansion,no cooling occurs because the molecules
A
Are above the inversion temperature
B
Exert no attractive force on each other
C
Do work equal to loss in kinetic energy
D
Collide without loss of energy
Solution
(B) In an ideal gas,there are no intermolecular forces of attraction between the particles.
During unrestrained expansion (expansion against vacuum),no work is done by the gas $(w = 0)$.
Since there are no attractive forces to overcome,no internal energy is consumed to separate the molecules,and therefore,the temperature of the ideal gas remains constant.
During unrestrained expansion (expansion against vacuum),no work is done by the gas $(w = 0)$.
Since there are no attractive forces to overcome,no internal energy is consumed to separate the molecules,and therefore,the temperature of the ideal gas remains constant.
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7
ChemistryDifficultMCQIIT JEE · 1984
Pure ammonia is placed in a vessel at a temperature where its dissociation constant $(\alpha)$ is appreciable. At equilibrium,
A
$K_p$ does not change significantly with pressure
B
$\alpha$ does not change with pressure
C
Concentration of $NH_3$ does not change with pressure
D
Concentration of $H_2$ is less than that of $N_2$
Solution
(A) The dissociation reaction of ammonia is: $2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$.
$K_p$ is the equilibrium constant,which depends only on temperature for a given reaction.
Therefore,$K_p$ does not change with pressure at a constant temperature.
Thus,option $A$ is correct.
$K_p$ is the equilibrium constant,which depends only on temperature for a given reaction.
Therefore,$K_p$ does not change with pressure at a constant temperature.
Thus,option $A$ is correct.
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8
ChemistryEasyMCQIIT JEE · 1984
The following gaseous reaction is taking place in a vessel: $C_2H_4(g) + H_2(g) \rightleftharpoons C_2H_6(g)$; $\Delta H = -32.7 \ kcal$. Which of the following will increase the equilibrium concentration of $C_2H_6$?
A
Increase of temperature
B
By reducing temperature
C
By removing some hydrogen
D
By adding some $C_2H_6$
Solution
(B) The given reaction is $C_2H_4(g) + H_2(g) \rightleftharpoons C_2H_6(g)$ with $\Delta H = -32.7 \ kcal$.
Since $\Delta H$ is negative,the reaction is exothermic.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature shifts the equilibrium in the forward direction.
Therefore,reducing the temperature will increase the yield and equilibrium concentration of the product $C_2H_6$.
Since $\Delta H$ is negative,the reaction is exothermic.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature shifts the equilibrium in the forward direction.
Therefore,reducing the temperature will increase the yield and equilibrium concentration of the product $C_2H_6$.
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9
ChemistryDifficultMCQIIT JEE · 1984
$A$ certain buffer solution contains equal concentrations of $3.9 \times 10^{-5} \ M$ $X^{-}$ and $HX$. The $K_b$ for $X^{-}$ is $10^{-10}$. The $pH$ of the buffer is:
A
$4$
B
$7$
C
$10$
D
$14$
Solution
(A) For the base $X^{-}$,the equilibrium is: $X^{-} + H_2O \rightleftharpoons OH^{-} + HX$.
Given $K_b = 10^{-10}$.
We know that $K_a \times K_b = K_w = 10^{-14}$.
Therefore,$K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4}$.
For a buffer solution,the Henderson-Hasselbalch equation is: $pH = pK_a + \log\frac{[salt]}{[acid]}$.
Since the concentrations of $X^{-}$ (salt) and $HX$ (acid) are equal,$[X^{-}] = [HX]$.
Thus,$pH = pK_a + \log(1) = pK_a$.
$pH = -\log(K_a) = -\log(10^{-4}) = 4$.
Given $K_b = 10^{-10}$.
We know that $K_a \times K_b = K_w = 10^{-14}$.
Therefore,$K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4}$.
For a buffer solution,the Henderson-Hasselbalch equation is: $pH = pK_a + \log\frac{[salt]}{[acid]}$.
Since the concentrations of $X^{-}$ (salt) and $HX$ (acid) are equal,$[X^{-}] = [HX]$.
Thus,$pH = pK_a + \log(1) = pK_a$.
$pH = -\log(K_a) = -\log(10^{-4}) = 4$.
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10
ChemistryEasyMCQIIT JEE · 1984
Hydrogen gas will not reduce
A
Heated cupric oxide
B
Heated ferric oxide
C
Heated stannic oxide
D
Heated aluminium oxide
Solution
(D) Aluminium oxide $(Al_2O_3)$ cannot be reduced by hydrogen even under very hot conditions because $Al$ is more reactive than $H$.
Hydrogen can only reduce metal oxides of metals that are less reactive than hydrogen in the metal reactivity series,such as $Cu$,$Fe$,and $Sn$.
Hydrogen can only reduce metal oxides of metals that are less reactive than hydrogen in the metal reactivity series,such as $Cu$,$Fe$,and $Sn$.
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11
ChemistryMediumMCQIIT JEE · 1984
The $IUPAC$ name of $(CH_3)_3C-CH=CH_2$ is
A
$3,3,3-$trimethylprop$-1-$ene
B
$1,1,1-$trimethylprop$-2-$ene
C
$3,3-$dimethylbut$-1-$ene
D
$2,2-$dimethylbut$-3-$ene
Solution
(C) The structure is $(CH_3)_3C-CH=CH_2$.
$1$. Identify the longest carbon chain containing the double bond: The longest chain has $4$ carbon atoms,so the parent alkane is butane,and with the double bond,it is but$-1-$ene.
$2$. Number the chain starting from the end closer to the double bond: $C_1=CH_2$,$C_2=CH$,$C_3=C(CH_3)_2$,$C_4=CH_3$.
$3$. Identify substituents: There are two methyl groups at the $C_3$ position.
$4$. Combine the parts: $3,3-$dimethylbut$-1-$ene.
$1$. Identify the longest carbon chain containing the double bond: The longest chain has $4$ carbon atoms,so the parent alkane is butane,and with the double bond,it is but$-1-$ene.
$2$. Number the chain starting from the end closer to the double bond: $C_1=CH_2$,$C_2=CH$,$C_3=C(CH_3)_2$,$C_4=CH_3$.
$3$. Identify substituents: There are two methyl groups at the $C_3$ position.
$4$. Combine the parts: $3,3-$dimethylbut$-1-$ene.
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12
ChemistryMediumMCQIIT JEE · 1984
Resonance structures of a molecule do not have:
A
Identical arrangement of atoms
B
Nearly the same energy content
C
The same number of paired electrons
D
Identical bonding
Solution
(D) Resonance structures are different Lewis structures for the same molecule that differ only in the distribution of electrons.
$(1)$ They must have the same arrangement of atoms.
$(2)$ They should have nearly the same energy content.
$(3)$ They must have the same number of paired and unpaired electrons.
$(4)$ They do not have identical bonding because the position of $\pi$-electrons or lone pairs changes between structures.
Therefore,the correct answer is $(d)$.
$(1)$ They must have the same arrangement of atoms.
$(2)$ They should have nearly the same energy content.
$(3)$ They must have the same number of paired and unpaired electrons.
$(4)$ They do not have identical bonding because the position of $\pi$-electrons or lone pairs changes between structures.
Therefore,the correct answer is $(d)$.
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13
ChemistryMediumMCQIIT JEE · 1984
An isotone of $_{32}^{76}Ge$ is (one or more are correct)
A
$_{32}^{77}Ge$
B
$_{33}^{77}As$
C
$_{34}^{78}Se$
D
Both $(b)$ and $(c)$
Solution
(D) Isotones are atoms that have the same number of neutrons.
For $_{32}^{76}Ge$,the number of neutrons is $76 - 32 = 44$.
For option $(a)$,$_{32}^{77}Ge$ has $77 - 32 = 45$ neutrons.
For option $(b)$,$_{33}^{77}As$ has $77 - 33 = 44$ neutrons.
For option $(c)$,$_{34}^{78}Se$ has $78 - 34 = 44$ neutrons.
Since both $(b)$ and $(c)$ have $44$ neutrons,they are both isotones of $_{32}^{76}Ge$.
For $_{32}^{76}Ge$,the number of neutrons is $76 - 32 = 44$.
For option $(a)$,$_{32}^{77}Ge$ has $77 - 32 = 45$ neutrons.
For option $(b)$,$_{33}^{77}As$ has $77 - 33 = 44$ neutrons.
For option $(c)$,$_{34}^{78}Se$ has $78 - 34 = 44$ neutrons.
Since both $(b)$ and $(c)$ have $44$ neutrons,they are both isotones of $_{32}^{76}Ge$.
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14
ChemistryMediumMCQIIT JEE · 1984
Commercially,methanol is prepared by:
A
Reduction of $CO$ in the presence of $ZnO-Cr_2O_3$
B
Methane reacting with water vapours at $900 \ ^\circ C$ in the presence of $Ni$ catalyst
C
Reduction of $HCHO$ by $LiAlH_4$
D
Reduction of $HCHO$ by aqueous $NaOH$
Solution
(A) Commercially,methanol is prepared by the catalytic hydrogenation of carbon monoxide.
The reaction is: $CO(g) + 2H_2(g) \xrightarrow{ZnO-Cr_2O_3} CH_3OH(l)$
This process occurs at high temperature and pressure in the presence of a catalyst mixture of zinc oxide and chromium oxide.
The reaction is: $CO(g) + 2H_2(g) \xrightarrow{ZnO-Cr_2O_3} CH_3OH(l)$
This process occurs at high temperature and pressure in the presence of a catalyst mixture of zinc oxide and chromium oxide.
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15
ChemistryEasyMCQIIT JEE · 1984
Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon?
A
$3s$
B
$2p$
C
$2s$
D
$1s$
Solution
(D) The $1s$ orbital is the ground state of the hydrogen atom,which has the lowest energy.
An electron in the $1s$ orbital can absorb a photon to transition to a higher energy level (excited state).
However,since there is no energy level lower than $1s$,an electron in the $1s$ orbital cannot emit a photon to transition to a lower state.
An electron in the $1s$ orbital can absorb a photon to transition to a higher energy level (excited state).
However,since there is no energy level lower than $1s$,an electron in the $1s$ orbital cannot emit a photon to transition to a lower state.
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16
ChemistryEasyMCQIIT JEE · 1984
Which of the following is not favourable for $SO_3$ formation in the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}; \Delta H = -45.0 \ kcal$?
A
High pressure
B
High temperature
C
Decreasing $SO_3$ concentration
D
Increasing reactant concentration
Solution
(B) The given reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ with $\Delta H = -45.0 \ kcal$.
Since the reaction is exothermic $(\Delta H < 0)$,according to Le Chatelier's principle,an increase in temperature will shift the equilibrium in the backward direction to absorb the excess heat.
Therefore,high temperature is not favourable for the formation of $SO_3$.
Since the reaction is exothermic $(\Delta H < 0)$,according to Le Chatelier's principle,an increase in temperature will shift the equilibrium in the backward direction to absorb the excess heat.
Therefore,high temperature is not favourable for the formation of $SO_3$.
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17
ChemistryMediumMCQIIT JEE · 1984
Which electron level would allow the hydrogen atom to absorb a photon but not emit a photon?
A
$3s$
B
$2p$
C
$2s$
D
$1s$
Solution
(D) The $1s$ electronic level allows the hydrogen atom to absorb a photon but not to emit one.
On absorption of radiation,the electron is transferred from the $1s$ level to a higher energy level,such as $2s$ or $3s$.
However,to emit a photon,an electron must transition from a higher energy level to a lower energy level.
Since $1s$ is the ground state (the lowest energy level),the electron cannot transition to a lower energy level.
Therefore,emission is not possible from the $1s$ state.
On absorption of radiation,the electron is transferred from the $1s$ level to a higher energy level,such as $2s$ or $3s$.
However,to emit a photon,an electron must transition from a higher energy level to a lower energy level.
Since $1s$ is the ground state (the lowest energy level),the electron cannot transition to a lower energy level.
Therefore,emission is not possible from the $1s$ state.
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18
ChemistryMCQIIT JEE · 1984
Pure ammonia is placed in a vessel at a temperature where its dissociation constant $(\alpha)$ is appreciable. At equilibrium,
A
$K_p$ does not change significantly with pressure.
B
$\alpha$ does not change with pressure.
C
concentration of $NH_3$ does not change with pressure.
D
concentration of hydrogen is less than that of nitrogen.
Solution
(A) The dissociation reaction is $2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$.
$K_p$ is a constant that depends only on temperature,not on pressure.
According to Le Chatelier's principle,as pressure increases,the equilibrium shifts towards the side with fewer moles of gas (left side),so the degree of dissociation $(\alpha)$ decreases.
From the stoichiometry of the reaction,for every $1$ mole of $N_2$ produced,$3$ moles of $H_2$ are produced,therefore the concentration of hydrogen is greater than that of nitrogen $([H_2] > [N_2])$.
Thus,$K_p$ remains constant with pressure.
$K_p$ is a constant that depends only on temperature,not on pressure.
According to Le Chatelier's principle,as pressure increases,the equilibrium shifts towards the side with fewer moles of gas (left side),so the degree of dissociation $(\alpha)$ decreases.
From the stoichiometry of the reaction,for every $1$ mole of $N_2$ produced,$3$ moles of $H_2$ are produced,therefore the concentration of hydrogen is greater than that of nitrogen $([H_2] > [N_2])$.
Thus,$K_p$ remains constant with pressure.
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19
ChemistryMCQIIT JEE · 1984
$A$ transverse wave is described by the equation $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to four times the wave velocity if
A
$\lambda = \frac{\pi y_0}{4}$
B
$\lambda = \frac{\pi y_0}{2}$
C
$\lambda = \pi y_0$
D
$\lambda = 2\pi y_0$
Solution
(B) The equation of the wave is $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$.
The particle velocity $v_p$ is given by the derivative of displacement with respect to time: $v_p = \frac{\partial y}{\partial t} = y_0 (2\pi f) \cos 2\pi \left( ft - \frac{x}{\lambda} \right)$.
The maximum particle velocity is $V_{\max} = 2\pi f y_0$.
The wave velocity $V$ is given by $V = f\lambda$.
According to the problem,$V_{\max} = 4V$.
Substituting the expressions,we get: $2\pi f y_0 = 4(f\lambda)$.
Dividing both sides by $f$,we get: $2\pi y_0 = 4\lambda$.
Therefore,$\lambda = \frac{2\pi y_0}{4} = \frac{\pi y_0}{2}$.
The particle velocity $v_p$ is given by the derivative of displacement with respect to time: $v_p = \frac{\partial y}{\partial t} = y_0 (2\pi f) \cos 2\pi \left( ft - \frac{x}{\lambda} \right)$.
The maximum particle velocity is $V_{\max} = 2\pi f y_0$.
The wave velocity $V$ is given by $V = f\lambda$.
According to the problem,$V_{\max} = 4V$.
Substituting the expressions,we get: $2\pi f y_0 = 4(f\lambda)$.
Dividing both sides by $f$,we get: $2\pi y_0 = 4\lambda$.
Therefore,$\lambda = \frac{2\pi y_0}{4} = \frac{\pi y_0}{2}$.
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20
ChemistryMCQIIT JEE · 1984
$A$ transverse wave is described by the equation $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to four times the wave velocity if:
A
$\lambda = \frac{\pi y_0}{4}$
B
$\lambda = \frac{\pi y_0}{2}$
C
$\lambda = \pi y_0$
D
$\lambda = 2\pi y_0$
Solution
(B) The given equation of the transverse wave is $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$.
The particle velocity $v_p$ is given by the derivative of displacement with respect to time:
$v_p = \frac{\partial y}{\partial t} = y_0 \cdot 2\pi f \cos 2\pi \left( ft - \frac{x}{\lambda} \right)$.
The maximum particle velocity is $v_{\max} = 2\pi f y_0$.
The wave velocity $v_w$ is given by $v_w = f\lambda$.
According to the problem,$v_{\max} = 4 v_w$.
Substituting the expressions:
$2\pi f y_0 = 4 (f\lambda)$.
Dividing both sides by $2f$:
$\pi y_0 = 2\lambda$.
Therefore,$\lambda = \frac{\pi y_0}{2}$.
The particle velocity $v_p$ is given by the derivative of displacement with respect to time:
$v_p = \frac{\partial y}{\partial t} = y_0 \cdot 2\pi f \cos 2\pi \left( ft - \frac{x}{\lambda} \right)$.
The maximum particle velocity is $v_{\max} = 2\pi f y_0$.
The wave velocity $v_w$ is given by $v_w = f\lambda$.
According to the problem,$v_{\max} = 4 v_w$.
Substituting the expressions:
$2\pi f y_0 = 4 (f\lambda)$.
Dividing both sides by $2f$:
$\pi y_0 = 2\lambda$.
Therefore,$\lambda = \frac{\pi y_0}{2}$.
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21
ChemistryMCQIIT JEE · 1984
$A$ transverse wave is described by the equation $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to four times the wave velocity if:
A
$\lambda = \frac{\pi y_0}{4}$
B
$\lambda = \frac{\pi y_0}{2}$
C
$\lambda = \pi y_0$
D
$\lambda = 2\pi y_0$
Solution
(B) The given wave equation is $y = y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$.
The particle velocity $v_p$ is given by the partial derivative of $y$ with respect to time $t$:
$v_p = \frac{\partial y}{\partial t} = y_0 \cdot 2\pi f \cos 2\pi \left( ft - \frac{x}{\lambda} \right)$.
The maximum particle velocity is $(v_p)_{\max} = 2\pi f y_0$.
The wave velocity $v_w$ is given by $v_w = f \lambda$.
According to the problem,the maximum particle velocity is four times the wave velocity:
$(v_p)_{\max} = 4 v_w$.
Substituting the expressions:
$2\pi f y_0 = 4 f \lambda$.
Dividing both sides by $4f$:
$\lambda = \frac{2\pi f y_0}{4f} = \frac{\pi y_0}{2}$.
The particle velocity $v_p$ is given by the partial derivative of $y$ with respect to time $t$:
$v_p = \frac{\partial y}{\partial t} = y_0 \cdot 2\pi f \cos 2\pi \left( ft - \frac{x}{\lambda} \right)$.
The maximum particle velocity is $(v_p)_{\max} = 2\pi f y_0$.
The wave velocity $v_w$ is given by $v_w = f \lambda$.
According to the problem,the maximum particle velocity is four times the wave velocity:
$(v_p)_{\max} = 4 v_w$.
Substituting the expressions:
$2\pi f y_0 = 4 f \lambda$.
Dividing both sides by $4f$:
$\lambda = \frac{2\pi f y_0}{4f} = \frac{\pi y_0}{2}$.
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22
ChemistryEasyMCQIIT JEE · 1984
To deposit one $gm$ equivalent of an element at an electrode,the quantity of electricity needed is
A
$1 \, \text{ampere}$
B
$96000 \, \text{amperes}$
C
$96500 \, \text{farads}$
D
$96500 \, \text{coulombs}$
Solution
(D) According to Faraday's laws of electrolysis,the amount of chemical change produced by an electric current is proportional to the quantity of electricity passed through the electrolyte.
One gram equivalent of any substance is deposited by the passage of $1 \, \text{Faraday}$ of electricity.
$1 \, \text{Faraday} = 96500 \, \text{coulombs}$.
Therefore,$96500 \, \text{coulombs}$ of charge is required to deposit one gram equivalent of an element at an electrode.
One gram equivalent of any substance is deposited by the passage of $1 \, \text{Faraday}$ of electricity.
$1 \, \text{Faraday} = 96500 \, \text{coulombs}$.
Therefore,$96500 \, \text{coulombs}$ of charge is required to deposit one gram equivalent of an element at an electrode.
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23
ChemistryMediumMCQIIT JEE · 1984
The radiations from a naturally occurring radio element,as seen after deflection in a magnetic field in one direction,are
A
Definitely $\alpha$-rays
B
Definitely $\beta$-rays
C
Both $\alpha$ and $\beta$-rays
D
Either $\alpha$ or $\beta$-rays
Solution
(D) The radiations emitted from a radioactive element consist of $\alpha$,$\beta$,and $\gamma$ rays.
$\gamma$-rays are neutral and do not deflect in a magnetic field.
Both $\alpha$-particles (positively charged) and $\beta$-particles (negatively charged) are deflected by a magnetic field,but they deflect in opposite directions.
If a radiation is observed to deflect in a specific direction,it could be either an $\alpha$-ray or a $\beta$-ray,depending on the charge of the particle and the orientation of the magnetic field.
Therefore,the correct option is $D$.
$\gamma$-rays are neutral and do not deflect in a magnetic field.
Both $\alpha$-particles (positively charged) and $\beta$-particles (negatively charged) are deflected by a magnetic field,but they deflect in opposite directions.
If a radiation is observed to deflect in a specific direction,it could be either an $\alpha$-ray or a $\beta$-ray,depending on the charge of the particle and the orientation of the magnetic field.
Therefore,the correct option is $D$.
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24
ChemistryMediumMCQIIT JEE · 1984
The electric charge required for the electrolytic decomposition of one gram equivalent of a substance is
A
One ampere per second
B
$96500 \, C$
C
One ampere for one hour
D
Charge on one mole of electrons
Solution
(B) According to Faraday's laws of electrolysis,the amount of electricity required to deposit or liberate one gram equivalent of a substance is equal to one Faraday.
One Faraday is equal to approximately $96500 \, C$,which is the charge carried by one mole of electrons.
One Faraday is equal to approximately $96500 \, C$,which is the charge carried by one mole of electrons.
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25
ChemistryDifficultMCQIIT JEE · 1984
$A$ solution containing $1 \ mol \ L^{-1}$ of each $Cu(NO_3)_2, AgNO_3, Hg_2(NO_3)_2$ and $Mg(NO_3)_2$ is being electrolysed using inert electrodes. The standard reduction potentials are $E^0_{Ag^{+}/Ag} = +0.80 \ V, E^0_{Hg_2^{2+}/Hg} = +0.79 \ V, E^0_{Cu^{2+}/Cu} = +0.34 \ V, E^0_{Mg^{2+}/Mg} = -2.37 \ V$. With increasing voltage,the sequence of deposition of metals on the cathode will be:
A
$Ag, Hg, Cu, Mg$
B
$Mg, Cu, Hg, Ag$
C
$Ag, Hg, Cu$
D
$Cu, Hg, Ag$
Solution
(C) During electrolysis,the cation with the highest standard reduction potential is reduced first at the cathode.
Comparing the given reduction potentials: $E^0_{Ag^{+}/Ag} (+0.80 \ V) > E^0_{Hg_2^{2+}/Hg} (+0.79 \ V) > E^0_{Cu^{2+}/Cu} (+0.34 \ V) > E^0_{Mg^{2+}/Mg} (-2.37 \ V)$.
Therefore,the order of deposition is $Ag$,then $Hg$,then $Cu$.
$Mg^{2+}$ ions are not reduced because the reduction of water $(2H_2O + 2e^- \rightarrow H_2 + 2OH^-)$ occurs at a more positive potential than the reduction of $Mg^{2+}$ in an aqueous solution.
Thus,the sequence of deposition is $Ag, Hg, Cu$.
Comparing the given reduction potentials: $E^0_{Ag^{+}/Ag} (+0.80 \ V) > E^0_{Hg_2^{2+}/Hg} (+0.79 \ V) > E^0_{Cu^{2+}/Cu} (+0.34 \ V) > E^0_{Mg^{2+}/Mg} (-2.37 \ V)$.
Therefore,the order of deposition is $Ag$,then $Hg$,then $Cu$.
$Mg^{2+}$ ions are not reduced because the reduction of water $(2H_2O + 2e^- \rightarrow H_2 + 2OH^-)$ occurs at a more positive potential than the reduction of $Mg^{2+}$ in an aqueous solution.
Thus,the sequence of deposition is $Ag, Hg, Cu$.
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26
ChemistryDifficultMCQIIT JEE · 1984
Which compound gives a yellow precipitate with iodine and alkali?
A
$2$-hydroxyethane
B
Acetophenone
C
Methyl acetone
D
Acetamide
Solution
(B) The reaction of a compound with iodine $(I_2)$ and alkali $(NaOH)$ is known as the iodoform test.
Compounds containing the $CH_3CO-$ group or $CH_3CH(OH)-$ group give a yellow precipitate of iodoform $(CHI_3)$.
Acetophenone $(C_6H_5COCH_3)$ contains the $CH_3CO-$ group.
The reaction is:
$C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 (\text{yellow ppt.}) + C_6H_5COONa + 3NaI + 3H_2O$.
Compounds containing the $CH_3CO-$ group or $CH_3CH(OH)-$ group give a yellow precipitate of iodoform $(CHI_3)$.
Acetophenone $(C_6H_5COCH_3)$ contains the $CH_3CO-$ group.
The reaction is:
$C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 (\text{yellow ppt.}) + C_6H_5COONa + 3NaI + 3H_2O$.
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27
ChemistryAdvancedMCQIIT JEE · 1984
$C_6H_5Cl$ is prepared from aniline by:
A
$HCl$
B
$Cu_2Cl_2$
C
$Cl_2$ in the presence of anhydrous $AlCl_3$
D
Treatment with $HNO_2$ followed by heating with $Cu_2Cl_2$
Solution
(D) The preparation of chlorobenzene $(C_6H_5Cl)$ from aniline involves two main steps:
$1$. Diazotization: Aniline reacts with $HNO_2$ (prepared in situ from $NaNO_2 + HCl$) at $0-5 \ ^\circ C$ to form benzene diazonium chloride.
$2$. Sandmeyer reaction: The benzene diazonium chloride is then treated with cuprous chloride $(Cu_2Cl_2)$ in the presence of $HCl$ to yield chlorobenzene,with the evolution of nitrogen gas $(N_2)$.
$1$. Diazotization: Aniline reacts with $HNO_2$ (prepared in situ from $NaNO_2 + HCl$) at $0-5 \ ^\circ C$ to form benzene diazonium chloride.
$2$. Sandmeyer reaction: The benzene diazonium chloride is then treated with cuprous chloride $(Cu_2Cl_2)$ in the presence of $HCl$ to yield chlorobenzene,with the evolution of nitrogen gas $(N_2)$.
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28
ChemistryDifficultMCQIIT JEE · 1984
Carbylamine test is performed by heating alcoholic $KOH$ with:
A
Chloroform and silver powder
B
Trihalogen methane and primary amine
C
Alkyl halide and primary amine
D
Alkyl cyanide and primary amine
Solution
(B) The Carbylamine reaction is a diagnostic test for primary amines.
In this reaction,a primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$.
The general reaction is: $R-NH_2 + CHCl_3 + 3KOH(alc.) \to R-NC + 3KCl + 3H_2O$.
Since chloroform is a trihalogen methane,option $B$ is the correct description.
In this reaction,a primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$.
The general reaction is: $R-NH_2 + CHCl_3 + 3KOH(alc.) \to R-NC + 3KCl + 3H_2O$.
Since chloroform is a trihalogen methane,option $B$ is the correct description.
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29
ChemistryMediumMCQIIT JEE · 1984
Which of the following compounds will react with ethanolic $KCN$?
A
Ethane
B
Acetyl chloride
C
Chlorobenzene
D
Benzaldehyde
Solution
(D) Benzaldehyde undergoes benzoin condensation in the presence of ethanolic $KCN$ to form benzoin.
$2C_6H_5CHO \xrightarrow{Alc. KCN} C_6H_5CH(OH)COC_6H_5$ (Benzoin).
Therefore,the correct option is $(D)$.
$2C_6H_5CHO \xrightarrow{Alc. KCN} C_6H_5CH(OH)COC_6H_5$ (Benzoin).
Therefore,the correct option is $(D)$.
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30
ChemistryMediumMCQIIT JEE · 1984
$A$ catalyst:
A
Increases the average kinetic energy of reacting molecules
B
Increases the activation energy
C
Alters the reaction mechanism
D
Increases the frequency of collisions of reacting species
Solution
(C) catalyst provides an alternative pathway for the reaction with lower activation energy. By lowering the activation energy,a larger fraction of molecules can cross the energy barrier,which effectively increases the rate of the reaction. While the frequency of collisions is determined by temperature and concentration,the catalyst's primary role is to alter the reaction mechanism to a lower energy path. However,in the context of standard multiple-choice questions regarding the effect of catalysts on reaction rates,option $C$ is the most scientifically accurate description of how a catalyst functions.
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