2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Morning Shift
A 0.5 percent solution of potassium chloride was found to freeze at $-$0.24$^\circ$C. The percentage dissociation of potassium chloride is ______________. (Nearest integer)
(Molal depression constant for water is 1.80 K kg mol$-$1 and molar mass of KCl is 74.6 g mol$-$1)
Correct Answer: 98
Explanation:
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{b}} \mathrm{m}$
$ \begin{aligned} &\mathrm{i}=\frac{0.24 \times 99.5 \times 74.6}{1.80 \times 0.5 \times 1000} \\\\ &=1.98 \\\\ &\alpha=\frac{\mathrm{i}-1}{\mathrm{n}-1}=\frac{0.98}{1}=0.98 = 98 \,\% \end{aligned} $
$ \begin{aligned} &\mathrm{i}=\frac{0.24 \times 99.5 \times 74.6}{1.80 \times 0.5 \times 1000} \\\\ &=1.98 \\\\ &\alpha=\frac{\mathrm{i}-1}{\mathrm{n}-1}=\frac{0.98}{1}=0.98 = 98 \,\% \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 24th June Evening Shift
A company dissolves 'x' amount of CO2 at 298 K in 1 litre of water to prepare soda water. X = __________ $\times$ 10$-$3 g. (nearest integer)
(Given : partial pressure of CO2 at 298 K = 0.835 bar.
Henry's law constant for CO2 at 298 K = 1.67 kbar.
Atomic mass of H, C and O is 1, 12, and 6 g mol$-$1, respectively)
Correct Answer: 1221
Explanation:
According to Henry's law, partial pressure of a gas is given by
$P_{g}=\left(K_{H}\right) X_{g}$
where $X_{g}$ is mole fraction of gas in solution
$0.835=1.67 \times 10^{3}\left(\mathrm{X}_{\mathrm{CO}_{2}}\right)$
$\mathrm{X}_{\mathrm{CO}_{2}}=5 \times 10^{-4}$
Mass of $\mathrm{CO}_{2}$ in $1 \mathrm{~L}$ water $=1221 \times 10^{-3} \mathrm{~g}$
$P_{g}=\left(K_{H}\right) X_{g}$
where $X_{g}$ is mole fraction of gas in solution
$0.835=1.67 \times 10^{3}\left(\mathrm{X}_{\mathrm{CO}_{2}}\right)$
$\mathrm{X}_{\mathrm{CO}_{2}}=5 \times 10^{-4}$
Mass of $\mathrm{CO}_{2}$ in $1 \mathrm{~L}$ water $=1221 \times 10^{-3} \mathrm{~g}$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 24th June Morning Shift
The osmotic pressure of blood is 7.47 bar at 300 K. To inject glucose to a patient intravenously, it has to be isotonic with blood. The concentration of glucose solution in gL$-$1 is _____________.
(Molar mass of glucose = 180 g mol$-$1, R = 0.083 L bar K$-$1 mol$-$1) (Nearest integer)
Correct Answer: 54
Explanation:
$7.47=\mathrm{C} \times 0.083 \times 300$
$(\pi=\mathrm{CRT})$
(Where C represents the concentration of glucose solution and $\pi$ represents osmotic pressure)
$\mathrm{C}=\frac{7.47}{0.083 \times 300}\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)$
which in $\mathrm{gm} / \mathrm{L}=\frac{7.47}{0.083 \times 300} \times 180$
$ =54 \mathrm{gm} / \mathrm{l} $
$(\pi=\mathrm{CRT})$
(Where C represents the concentration of glucose solution and $\pi$ represents osmotic pressure)
$\mathrm{C}=\frac{7.47}{0.083 \times 300}\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)$
which in $\mathrm{gm} / \mathrm{L}=\frac{7.47}{0.083 \times 300} \times 180$
$ =54 \mathrm{gm} / \mathrm{l} $
2022
JEE Advanced
Numerical
JEE Advanced 2022 Paper 2 Online
An aqueous solution is prepared by dissolving $0.1 \mathrm{~mol}$ of an ionic salt in $1.8 \mathrm{~kg}$ of water at $35^{\circ} \mathrm{C}$. The salt remains $90 \%$ dissociated in the solution. The vapour pressure of the solution is $59.724 \mathrm{~mm}$ of Hg. Vapor pressure of water at $35{ }^{\circ} \mathrm{C}$ is $60.000 \mathrm{~mm}$ of $\mathrm{Hg}$. The number of ions present per formula unit of the ionic salt is _________.
Correct Answer: 5
Explanation:
Vapour pressure of solution $\left(\mathrm{P}_{\mathrm{A}}\right)$
$
=59.724 \mathrm{~mm} \text { of } \mathrm{Hg}
$
Vapour pressure of pure water $\left(\mathrm{P}_{\mathrm{A}}^{\circ}\right)$ $ =60.000 \mathrm{~mm} \text { of } \mathrm{Hg} $
Also, $0.1 \mathrm{~mol}$ of an ionic solid is dissolved in $1.8 \mathrm{~kg}$ of water and salt remains $90 \%$ dissocated in the solution.
So, total number of moles $=0.01+0.09 a$ of non-volatile particles.
Now, mass of water $=1.8 \mathrm{~kg}=1.8 \times 1.8 \times 1000 \mathrm{~g}$
Molar mass of water $=18 \mathrm{~g}$
Moles of water $=\frac{1.8 \times 1000}{18}=100$ moles
Using the colligative property, relative lowering in vapour pressure,
$ \frac{\mathrm{P}_{\mathrm{A}}^{\circ}-\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}^{\circ}}=x_{\mathrm{A}} $
$ \Rightarrow $ $\frac{60-59.724}{60} =\frac{0.01+0.09 a}{100} $
$ \Rightarrow $ $ \frac{0.276}{60} =\frac{0.01+0.09 a}{100} $
$ \Rightarrow $ $ \frac{27.6}{60} =0.01+0.09 a $
$ \Rightarrow $ $ 0.46 =0.01+0.09 a $
$ \Rightarrow $ $ 0.09 a =0.45 $
$ \Rightarrow $ $ a =\frac{0.45}{0.89} $
$ \Rightarrow $ $ a =5 $
So, the number of ions present per formula unit of the ionic salt is 5 .
Vapour pressure of pure water $\left(\mathrm{P}_{\mathrm{A}}^{\circ}\right)$ $ =60.000 \mathrm{~mm} \text { of } \mathrm{Hg} $
Also, $0.1 \mathrm{~mol}$ of an ionic solid is dissolved in $1.8 \mathrm{~kg}$ of water and salt remains $90 \%$ dissocated in the solution.
So, total number of moles $=0.01+0.09 a$ of non-volatile particles.
Now, mass of water $=1.8 \mathrm{~kg}=1.8 \times 1.8 \times 1000 \mathrm{~g}$
Molar mass of water $=18 \mathrm{~g}$
Moles of water $=\frac{1.8 \times 1000}{18}=100$ moles
Using the colligative property, relative lowering in vapour pressure,
$ \frac{\mathrm{P}_{\mathrm{A}}^{\circ}-\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}^{\circ}}=x_{\mathrm{A}} $
$ \Rightarrow $ $\frac{60-59.724}{60} =\frac{0.01+0.09 a}{100} $
$ \Rightarrow $ $ \frac{0.276}{60} =\frac{0.01+0.09 a}{100} $
$ \Rightarrow $ $ \frac{27.6}{60} =0.01+0.09 a $
$ \Rightarrow $ $ 0.46 =0.01+0.09 a $
$ \Rightarrow $ $ 0.09 a =0.45 $
$ \Rightarrow $ $ a =\frac{0.45}{0.89} $
$ \Rightarrow $ $ a =5 $
So, the number of ions present per formula unit of the ionic salt is 5 .
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?
A.
hydrazine
B.
glucose
C.
glycine
D.
KHSO4
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
Which one of the following 0.06 M aqueous solutions has lowest freezing point?
A.
Al2(SO4)3
B.
C6H12O6
C.
KI
D.
K2SO4
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
1.22 g of an organic acid is separately dissolved in 100 g of benzene (Kb = 2.6 K kg mol$-$1) and 100 g of acetone (Kb = 1.7 K kg mol$-$1). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by 0.17$^\circ$C. The increase in boiling point of solution in benzene in $^\circ$C is x $\times$ 10$-$2. The value of x is ______________. (Nearest integer) [Atomic mass : C = 12.0, H = 1.0, O =16.0]
Correct Answer: 13
Explanation:
With benzene as solvent
$\Delta$Tb = i Kb m
$\Delta$Tb = ${1 \over 2}$ $\times$ 2.6 $\times$ ${{1.22/{M_w}} \over {100/1000}}$ .... (1)
With Acetone as solvent
$\Delta$Tb = i Kb m
0.17 = 1 $\times$ 1.7 $\times$ ${{1.22/{M_w}} \over {100/1000}}$ ..... (2)
(1) / (2)
${{\Delta {T_b}} \over {0.17}} = {{{1 \over 2} \times 2.6 + {{1.22/{M_w}} \over {100/1000}}} \over {1 \times 1.7 \times {{1.22/{M_w}} \over {100/1000}}}}$
$\Delta$Tb = ${{0.26} \over 2}$
$\Delta$Tb = 13 $\times$ 10$-$2
$\Rightarrow$ x = 13
$\Delta$Tb = i Kb m
$\Delta$Tb = ${1 \over 2}$ $\times$ 2.6 $\times$ ${{1.22/{M_w}} \over {100/1000}}$ .... (1)
With Acetone as solvent
$\Delta$Tb = i Kb m
0.17 = 1 $\times$ 1.7 $\times$ ${{1.22/{M_w}} \over {100/1000}}$ ..... (2)
(1) / (2)
${{\Delta {T_b}} \over {0.17}} = {{{1 \over 2} \times 2.6 + {{1.22/{M_w}} \over {100/1000}}} \over {1 \times 1.7 \times {{1.22/{M_w}} \over {100/1000}}}}$
$\Delta$Tb = ${{0.26} \over 2}$
$\Delta$Tb = 13 $\times$ 10$-$2
$\Rightarrow$ x = 13
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is __________ K. (Nearest integer) [Given : Kf = 1.86 K kg mol$-$1; Density of water = 1.00 g cm$-$3; Freezing point of water = 273.15 K]
Correct Answer: 271
Explanation:
Molality = ${{\left( {{{40} \over {180}}} \right)mol} \over {0.2\,Kg}} = \left( {{{10} \over 9}} \right)$ molal
$ \Rightarrow \Delta {T_f} = {T_f} - {T_f}' = 1.86 \times {{10} \over 9}$
$ \Rightarrow {T_f}' = 273.15 - 1.86 \times {{10} \over 9}$
= 271.08 K
$ \simeq $ 271 K (nearest - integer)
$ \Rightarrow \Delta {T_f} = {T_f} - {T_f}' = 1.86 \times {{10} \over 9}$
$ \Rightarrow {T_f}' = 273.15 - 1.86 \times {{10} \over 9}$
= 271.08 K
$ \simeq $ 271 K (nearest - integer)
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $-$4$^\circ$C before freezing. The amount of ice (in g) that will be separated out is __________. (Nearest integer)
[Given : Kf(H2O) = 1.86 K kg mol$-$1]
[Given : Kf(H2O) = 1.86 K kg mol$-$1]
Correct Answer: 518
Explanation:
Let mass of water initially present = x gm
$\Rightarrow$ Mass of sucrose = (1000 $-$ x) gm
$\Rightarrow$ moles of sucrose = $\left( {{{1000 - x} \over {342}}} \right)$
$ \Rightarrow 0.75 = {{\left( {{{1000 - x} \over {342}}} \right)} \over {\left( {{x \over {1000}}} \right)}} \Rightarrow {x \over {1000}} = {{1000 - x} \over {342 \times 0.75}}$
$\Rightarrow$ 256.5x = 106 $-$ 1000x
$\Rightarrow$ x = 795.86 gm
$\Rightarrow$ moles of sucrose = 0.5969
New mass of H2O = a kg
$ \Rightarrow 4 = {{0.5969} \over a} \times 1.86 \Rightarrow $ a = 0.2775 kg
$\Rightarrow$ ice separated = (795.86 $-$ 277.5) = 518.3 gm
$\Rightarrow$ Mass of sucrose = (1000 $-$ x) gm
$\Rightarrow$ moles of sucrose = $\left( {{{1000 - x} \over {342}}} \right)$
$ \Rightarrow 0.75 = {{\left( {{{1000 - x} \over {342}}} \right)} \over {\left( {{x \over {1000}}} \right)}} \Rightarrow {x \over {1000}} = {{1000 - x} \over {342 \times 0.75}}$
$\Rightarrow$ 256.5x = 106 $-$ 1000x
$\Rightarrow$ x = 795.86 gm
$\Rightarrow$ moles of sucrose = 0.5969
New mass of H2O = a kg
$ \Rightarrow 4 = {{0.5969} \over a} \times 1.86 \Rightarrow $ a = 0.2775 kg
$\Rightarrow$ ice separated = (795.86 $-$ 277.5) = 518.3 gm
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ____________ K. (Nearest integer)
[Use : Molal Freezing point depression constant of water = 1.86 K kg mol$-$1]
Freezing Point of water = 273 K
Atomic masses : C : 12.0 u, O : 16.0 u, H : 1.0 u]
[Use : Molal Freezing point depression constant of water = 1.86 K kg mol$-$1]
Freezing Point of water = 273 K
Atomic masses : C : 12.0 u, O : 16.0 u, H : 1.0 u]
Correct Answer: 269
Explanation:
kf = 1.86 kg/mol
T$_f^o$ = 273 K
solvent : H2O(625 g)
Solute : 83 g of ethylene glycol
$\Rightarrow$ $\Delta$Tf = kf $\times$ m
$\Rightarrow$ $\left( {T_f^o - T_f^1} \right) = 1.86 \times {{83/62} \over {625/1000}}$
$ \Rightarrow 273 - T_f^1 = {{1.86 \times 83 \times 1000} \over {62 \times 625}} = {{154380} \over {38750}}$
$ \Rightarrow 273 - T_f^1 = 4$
$ \Rightarrow T_f^1 = 259$ K
T$_f^o$ = 273 K
solvent : H2O(625 g)
Solute : 83 g of ethylene glycol
$\Rightarrow$ $\Delta$Tf = kf $\times$ m
$\Rightarrow$ $\left( {T_f^o - T_f^1} \right) = 1.86 \times {{83/62} \over {625/1000}}$
$ \Rightarrow 273 - T_f^1 = {{1.86 \times 83 \times 1000} \over {62 \times 625}} = {{154380} \over {38750}}$
$ \Rightarrow 273 - T_f^1 = 4$
$ \Rightarrow T_f^1 = 259$ K
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Morning Shift
Of the following four aqueous solutions, total number of those solutions whose freezing point is lower than that of 0.10 M C2H5OH is __________ (Integer answer)
(i) 0.10 M Ba3(PO4)2
(ii) 0.10 M Na2SO4
(iii) 0.10 M KCl
(iv) 0.10 M Li3PO4
(i) 0.10 M Ba3(PO4)2
(ii) 0.10 M Na2SO4
(iii) 0.10 M KCl
(iv) 0.10 M Li3PO4
Correct Answer: 4
Explanation:
As 0.1 M C2H5OH is non-dissociative and rest all salt given are electrolyte so in each case effective molarity > 0.1 so each will have lower freezing point.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is __________ $\times$ 10$-$2.
(Round off to the nearest integer)
(Round off to the nearest integer)
Correct Answer: 125
Explanation:
Now, $i = {{final\,moles} \over {initial\,moles}} = {{0.25a + 0.5a + 0.5a} \over {0.5a + 0.5a}}$
= 1.25 = 125 $\times$ 10$-$2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 $\times$ 10$-$3 bar.
The molar mass of the biopolymer is _____________ $\times$ 104 g mol$-$1. (Round off to the Nearest Integer)
[Use : R = 0.083 L bar mol$-$1 K$-$1]
The molar mass of the biopolymer is _____________ $\times$ 104 g mol$-$1. (Round off to the Nearest Integer)
[Use : R = 0.083 L bar mol$-$1 K$-$1]
Correct Answer: 15
Explanation:
$\pi$ = CRT;
$\pi$ = osmotic pressure
C = molarity
T = Temperature of solution
let the molar mass be M gm/mol
2.42 $\times$ 10$-$3 bar = ${{\left( {{{1.46g} \over {Mgm/mol}}} \right)} \over {0.1l}} \times \left( {{{0.083l - bar} \over {mol - K}}} \right) \times (300K)$
$\Rightarrow$ M = 15.02 $\times$ 104 g/mol
$\pi$ = osmotic pressure
C = molarity
T = Temperature of solution
let the molar mass be M gm/mol
2.42 $\times$ 10$-$3 bar = ${{\left( {{{1.46g} \over {Mgm/mol}}} \right)} \over {0.1l}} \times \left( {{{0.083l - bar} \over {mol - K}}} \right) \times (300K)$
$\Rightarrow$ M = 15.02 $\times$ 104 g/mol
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
When 3.00 g of a substance 'X' is dissolved in 100 g of CCl4, it raises the boiling point by 0.60 K. The molar mass of the substance 'X' is ______________ g mol$-$1. (Nearest integer).
[Given Kb for CCl4 is 5.0 K kg mol$-$1]
[Given Kb for CCl4 is 5.0 K kg mol$-$1]
Correct Answer: 250
Explanation:
$\Delta$Tb = Kb $\times$ molality
0.60 = 5 $\times$ $\left( {{{3/M} \over {100/100}}} \right)$
M = 250
0.60 = 5 $\times$ $\left( {{{3/M} \over {100/100}}} \right)$
M = 250
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
CO2 gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO2 exerts a partial pressure of 0.835 bar then x m mol of CO2 would dissolve in 0.9 L of water. The value of x is ____________. (Nearest integer)
(Henry's law constant for CO2 at 298 K is 1.67 $\times$ 103 bar)
(Henry's law constant for CO2 at 298 K is 1.67 $\times$ 103 bar)
Correct Answer: 25
Explanation:
From Henry's law
Pgas = KH.Xgas
0.835 = 1.67 $\times$ 103 $\times$ ${{n(C{O_2})} \over {{{0.9 \times 1000} \over {18}}}}$
$ \Rightarrow $ n(CO2) = 0.025
Millimoles of CO2 = 0.025 $\times$ 1000 = 25
Pgas = KH.Xgas
0.835 = 1.67 $\times$ 103 $\times$ ${{n(C{O_2})} \over {{{0.9 \times 1000} \over {18}}}}$
$ \Rightarrow $ n(CO2) = 0.025
Millimoles of CO2 = 0.025 $\times$ 1000 = 25
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
The vapour pressures of A and B at 25$^\circ$C are 90 mm Hg and 15 mm Hg respectively. If A and B are mixed such that the mole fraction of A in the mixture is 0.6, then the mole fraction of B in the vapour phase is x $\times$ 10$-$1. The value of x is ___________. (Nearest integer)
Correct Answer: 1
Explanation:
xA = 0.6
PT = xAPAo + xBPBo
= 0.6 $\times$ 90 + 0.4 $\times$ 15
= 54 + 6 = 60
xAPAo = yAPT
0.6 $\times$ 90 = yA(60)
$\Rightarrow$ yA = 0.9
yB = 0.1 = 1 $\times$ 10$-$1
$\therefore$ x = 1
PT = xAPAo + xBPBo
= 0.6 $\times$ 90 + 0.4 $\times$ 15
= 54 + 6 = 60
xAPAo = yAPT
0.6 $\times$ 90 = yA(60)
$\Rightarrow$ yA = 0.9
yB = 0.1 = 1 $\times$ 10$-$1
$\therefore$ x = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
At 20$^\circ$ C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20$^\circ$ above an equimolar mixture of benzene and methyl benzene is _____________ $\times$ 10$-$2. (Nearest integer)
Correct Answer: 78
Explanation:
Vapour pressure of pure benzene, $p_A^o$ = 70 Torr
Vapour pressure of pure methyl benzene, $p_B^o$ = 20 Torr
This mixture is equimolar, so number of moles of benzene,
nA = number of moles of methyl benzene, nB
Mole fraction of benzene in vapour phase, yA = ${{{p_A}} \over {{p_T}}}$ .... (i)
where pA is pressure of benzene in mixture.
${p_A} = \mathop {p_A^o{\chi _A}}\limits_{Mole\,fraction} = p_A^o \times {{{n_A}} \over {{n_A} + {n_B}}} = p_A^o \times {{{n \over A}} \over {2{n \over A}}} = {{p_A^o} \over 2}$
pT is total pressure,
${p_T} = p_A^o{\chi _A} + p_B^o{\chi _B} = {p_A} + {p_B}$ (pressure of methyl benzene)
${p_T} = {{p_A^o} \over 2} + {{p_B^o} \over 2} = {{p_A^o + p_B^o} \over 2}$
Putting in above equation (i),
${y_A} = {{{{70} \over 2}} \over {{{(70 + 20)} \over 2}}} = {{70} \over {90}}$
yA = 0.78 = 78 $\times$ 10$-$2
Vapour pressure of pure methyl benzene, $p_B^o$ = 20 Torr
This mixture is equimolar, so number of moles of benzene,
nA = number of moles of methyl benzene, nB
Mole fraction of benzene in vapour phase, yA = ${{{p_A}} \over {{p_T}}}$ .... (i)
where pA is pressure of benzene in mixture.
${p_A} = \mathop {p_A^o{\chi _A}}\limits_{Mole\,fraction} = p_A^o \times {{{n_A}} \over {{n_A} + {n_B}}} = p_A^o \times {{{n \over A}} \over {2{n \over A}}} = {{p_A^o} \over 2}$
pT is total pressure,
${p_T} = p_A^o{\chi _A} + p_B^o{\chi _B} = {p_A} + {p_B}$ (pressure of methyl benzene)
${p_T} = {{p_A^o} \over 2} + {{p_B^o} \over 2} = {{p_A^o + p_B^o} \over 2}$
Putting in above equation (i),
${y_A} = {{{{70} \over 2}} \over {{{(70 + 20)} \over 2}}} = {{70} \over {90}}$
yA = 0.78 = 78 $\times$ 10$-$2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52$^\circ$C. The percentage association of A is __________. (Round off to the Nearest Integer).
[Use : Kb for water = 0.52 K kg mol$-$1 Boiling point of water = 100$^\circ$C]
[Use : Kb for water = 0.52 K kg mol$-$1 Boiling point of water = 100$^\circ$C]
Correct Answer: 100
Explanation:
$\Delta$Tb = Boiling point of the solution $-$ Boiling point of the pure solvent
= 100.52 $-$ 100
= 0.52
$ \therefore $ $\Delta$Tb = Kb (iM)
$ \Rightarrow $ 0.52 = i $\times$ 0.52 $\times$ 2
$ \Rightarrow $ i = ${1 \over 2}$
We know,
i = 1 + $\left( {{1 \over n} - 1} \right)\beta $
here, $\beta$ = degree of dimerization
n = number of particle associated
n = 2 for dimerization
$ \therefore $ ${1 \over 2} = 1 + \left( {{1 \over 2} - 1} \right)\beta $
$ \Rightarrow \beta = 1$
$ \therefore $ % association = 100
= 100.52 $-$ 100
= 0.52
$ \therefore $ $\Delta$Tb = Kb (iM)
$ \Rightarrow $ 0.52 = i $\times$ 0.52 $\times$ 2
$ \Rightarrow $ i = ${1 \over 2}$
We know,
i = 1 + $\left( {{1 \over n} - 1} \right)\beta $
here, $\beta$ = degree of dimerization
n = number of particle associated
n = 2 for dimerization
$ \therefore $ ${1 \over 2} = 1 + \left( {{1 \over 2} - 1} \right)\beta $
$ \Rightarrow \beta = 1$
$ \therefore $ % association = 100
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
2 molal solution of a weak acid HA has a freezing point of 3.885$^\circ$C. The degree of dissociation of this acid is ___________ $\times$ 10$-$3. (Round off to the Nearest Integer).
[Given : Molal depression constant of water = 1.85 K kg mol$-$1 Freezing point of pure water = 0$^\circ$ C]
[Given : Molal depression constant of water = 1.85 K kg mol$-$1 Freezing point of pure water = 0$^\circ$ C]
Correct Answer: 50
Explanation:
$\Delta$Tf = Kf (im)
$ \Rightarrow $ 3.885 = i $\times$ 1.85 $\times$ 2
$ \Rightarrow $ i = 1.05
Also, we know,
i = 1 + (n $-$ 1) $\alpha$
here n = number of particle obtained upon the dissociation of one particle.
$HA\rightleftharpoons H^{+}+A^{-} $
here from one particle HA we get two particle H+ and A$-$.
$ \therefore $ n = 2
So, i = 1 + (2 $-$ 1)$\alpha$
$ \Rightarrow $ 1.05 = 1 + $\alpha$
$ \Rightarrow $ $\alpha$ = 0.05 = 50 $\times$ 10$-$3
$ \Rightarrow $ 3.885 = i $\times$ 1.85 $\times$ 2
$ \Rightarrow $ i = 1.05
Also, we know,
i = 1 + (n $-$ 1) $\alpha$
here n = number of particle obtained upon the dissociation of one particle.
$HA\rightleftharpoons H^{+}+A^{-} $
here from one particle HA we get two particle H+ and A$-$.
$ \therefore $ n = 2
So, i = 1 + (2 $-$ 1)$\alpha$
$ \Rightarrow $ 1.05 = 1 + $\alpha$
$ \Rightarrow $ $\alpha$ = 0.05 = 50 $\times$ 10$-$3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
A 1 molal K4Fe(CN)6 solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is __________ u. (Round off to the Nearest Integer). [Density of water = 1.0 g cm$-$3 ]
Correct Answer: 85
Explanation:
Effective molality = 0.6 + 1.6 + 0.4 = 2.6 m
As elevation in boiling point is a colligative property which depends on the amount of solute. So, to have same boiling point, the molality of two solutions should be same.
Molality of non-electrolyte solution = molality of ${K_4}[Fe{(CN)_6}]$ = 2.6 m
Now, 18.1 weight per cent solution means 18.1 g solute is present in 100 g solution and hence, (100 $-$ 18.1) = 81.9 g water.
$Molality = {{(Mass\,of\,solute/Molar\,mass\,of\,solute)} \over {Mass\,of\,solvent\,in\,kg}} \times 1000$
Now, $2.6 = {{18.1/M} \over {81.9/1000}}$
where, M is the molar mass of non-electrolyte solute
Molar mass of solute, M = 85
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is __________ $\times$ 10$-$5 mol dm$-$3. (Round off to the Nearest Integer).
[Given : Henry's law constant = KH = 8.0 $\times$ 104 kPa for O2. Density of water with dissolved oxygen = 1.0 kg dm$-$3 ]
[Given : Henry's law constant = KH = 8.0 $\times$ 104 kPa for O2. Density of water with dissolved oxygen = 1.0 kg dm$-$3 ]
Correct Answer: 25
Explanation:
The oxygen dissolved in water has a partial pressure of 20 kPa in the vapor phase above the water. To find the molar solubility of oxygen in water, we use Henry's Law, which states:
$ \text{Partial pressure} (P_g) \propto \text{Solubility} $
This can be mathematically expressed as:
$ P_g = K_H \times \text{Solubility} $
Where:
$ P_g $ is the partial pressure of oxygen, given as 20 kPa.
$ K_H $ is Henry's law constant for $ \text{O}_2 $, provided as $ 8.0 \times 10^4 \, \text{kPa} $.
Substituting the given values into the equation:
$ 20 \times 10^3 = (8.0 \times 10^4) \times \text{Solubility} $
Solving for solubility:
$ \text{Solubility} = \frac{20 \times 10^3}{8.0 \times 10^4} $
$ \text{Solubility} = 0.25 \times 10^{-1} $
$ \text{Solubility} = 25 \times 10^{-5} \, \text{mol dm}^{-3} $
Thus, the molar solubility of oxygen in water rounds to 25 × 10-5 mol dm-3.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
At 363 K, the vapour pressure of A is 21 kPa and that of B is 18 kPa. One mole of A and 2 moles of B are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is ___________ kPa. (Round off to the Nearest Integer).
Correct Answer: 19
Explanation:
${X_A} = {1 \over {1 + 2}} = {1 \over 3}$
${X_B} = {2 \over 3}$
$P_A^o = 21kPa$
$P_B^o = 18kPa$
${P_{total}} = P_A^o{X_A} + P_B^o{X_B}$
$ = 21 \times {1 \over 3} + 18 \times {2 \over 3}$
$ = 7 + 12$
$ = 19 kPa$
${X_B} = {2 \over 3}$
$P_A^o = 21kPa$
$P_B^o = 18kPa$
${P_{total}} = P_A^o{X_A} + P_B^o{X_B}$
$ = 21 \times {1 \over 3} + 18 \times {2 \over 3}$
$ = 7 + 12$
$ = 19 kPa$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
AB2 is 10% dissociated in water to A2+ and B$-$. The boiling point of a 10.0 molal aqueous solution of AB2 is __________$^\circ$C. (Round off to the Nearest Integer).
[Given : Molal elevation constant of water Kb = 0.5 K kg mol$-$1 boiling point of pure water = 100$^\circ$C]
[Given : Molal elevation constant of water Kb = 0.5 K kg mol$-$1 boiling point of pure water = 100$^\circ$C]
Correct Answer: 106
Explanation:
AB2 $ \leftrightharpoons $ A+ + 2B$-$
$ \therefore $ For AB2, n = 3
i = 1 + (n $-$ 1)$\alpha$
= 1 + (3 $-$ 1) $\times$ 0.1
= 1.2
Now, $\Delta$Tb = Kb (im)
$ \Rightarrow $ Tb $-$ T$_b^o$ = 1.2 $\times$ 0.5 $\times$ 10
$ \Rightarrow $ Tb $-$ 100 = 6
$ \Rightarrow $ Tb = 106
$ \therefore $ For AB2, n = 3
i = 1 + (n $-$ 1)$\alpha$
= 1 + (3 $-$ 1) $\times$ 0.1
= 1.2
Now, $\Delta$Tb = Kb (im)
$ \Rightarrow $ Tb $-$ T$_b^o$ = 1.2 $\times$ 0.5 $\times$ 10
$ \Rightarrow $ Tb $-$ 100 = 6
$ \Rightarrow $ Tb = 106
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be $-$0.93$^\circ$C (Kf(H2O) = 1.86 K kg mol$-$1). The number (n) of benzoic acid molecules associated (assuming 100% association) is ___________.
Correct Answer: 2
Explanation:
$\underset{\text{Benzoic acid}}{n \mathrm{PhCOOH}} \stackrel{\text { Association }}{\longrightarrow}(\mathrm{PhCOOH})_n$
Assuming $100 \%$ association ( $\alpha=1$ ),
$ \Rightarrow i=1-\alpha\left(1-\frac{1}{n}\right)=\frac{1}{n}[\because \alpha+1] $
Now, $\Delta T_f=K_f \times m \times i$
$ 0-(0.93)=1.86 \times \frac{w_B \times 1000}{w_A \times M_B} \times \frac{1}{n} $
$\left[\because w_B=\right.$ mass of $\mathrm{PhCOOH}=12.2 \mathrm{~g}$
$w_A=$ mass of $\mathrm{H}_2 \mathrm{O}=100 \mathrm{~g}$
$M_B=$ molar mass of $\left.\mathrm{PhCOOH}\right]$
$=122 \mathrm{~g} \mathrm{~mol}^{-1}$
$=1.86 \times \frac{12.2 \times 1000}{100 \times 122} \times \frac{1}{n}$
$ \begin{aligned} \Rightarrow n &=\frac{1.86 \times 12.2 \times 1000}{0.93 \times 100 \times 122}=2 \end{aligned} $
$\therefore$ Number of benzoic acid molecules associated, $n=2$
Assuming $100 \%$ association ( $\alpha=1$ ),
$ \Rightarrow i=1-\alpha\left(1-\frac{1}{n}\right)=\frac{1}{n}[\because \alpha+1] $
Now, $\Delta T_f=K_f \times m \times i$
$ 0-(0.93)=1.86 \times \frac{w_B \times 1000}{w_A \times M_B} \times \frac{1}{n} $
$\left[\because w_B=\right.$ mass of $\mathrm{PhCOOH}=12.2 \mathrm{~g}$
$w_A=$ mass of $\mathrm{H}_2 \mathrm{O}=100 \mathrm{~g}$
$M_B=$ molar mass of $\left.\mathrm{PhCOOH}\right]$
$=122 \mathrm{~g} \mathrm{~mol}^{-1}$
$=1.86 \times \frac{12.2 \times 1000}{100 \times 122} \times \frac{1}{n}$
$ \begin{aligned} \Rightarrow n &=\frac{1.86 \times 12.2 \times 1000}{0.93 \times 100 \times 122}=2 \end{aligned} $
$\therefore$ Number of benzoic acid molecules associated, $n=2$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
224 mL of SO2(g) at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The non-volatile solute produced is dissolved in 36g of water. The lowering of vapour pressure of solution (assuming the solution in dilute) (P$_{({H_2}O)}^o$ $-$ 24 mm of Hg) is x $\times$ 10$-$2 mm of Hg, the value of x is ___________. (Integer answer)
Correct Answer: 18TO24
Explanation:
moles of SO2 = ${{224} \over {22400}}$ = 0.01
moles of NaOH = molarity × volume (in litre)
= 0.1 × 0.1
= 0.01 moles
The balanced equation is
SO2 + 2NaOH $ \to $ Na2SO3 + H2O
$ \therefore $ Here NaOH is limiting Reagent.
2 mole NaOH produces 1 mole Na2SO3
0.01 mole NaOH produces ${1 \over 2}$ $ \times $ 0.01 mole Na2SO3
$ \therefore $ Moles of Na2SO3 = 0.005 mole
Na2SO3 $ \to $ 2Na+ + SO32-
van’t Hoff factor (i) = 3
Moles of H2O = ${{36} \over {18}}$ = 2 moles
Accoding to Relative Lowering of Vapour :
${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}} + i{n_{N{a_2}C{O_3}}}}}$
[ ${{n_{N{a_2}C{O_3}}}}$ << ${{n_{{H_2}O}}}$ ]
${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}}}}$
$ \Rightarrow $ ${{24 - {P_S}} \over {24}} = {{3 \times 0.005} \over 2}$
$ \Rightarrow $ 24 – PS = 0.18
$ \Rightarrow $ PS = 23.82
Lowering in pressure ($\Delta $P) = 0.18 mm of Hg
= 18 × 10–2 mm of Hg
moles of NaOH = molarity × volume (in litre)
= 0.1 × 0.1
= 0.01 moles
The balanced equation is
SO2 + 2NaOH $ \to $ Na2SO3 + H2O
$ \therefore $ Here NaOH is limiting Reagent.
2 mole NaOH produces 1 mole Na2SO3
0.01 mole NaOH produces ${1 \over 2}$ $ \times $ 0.01 mole Na2SO3
$ \therefore $ Moles of Na2SO3 = 0.005 mole
Na2SO3 $ \to $ 2Na+ + SO32-
van’t Hoff factor (i) = 3
Moles of H2O = ${{36} \over {18}}$ = 2 moles
Accoding to Relative Lowering of Vapour :
${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}} + i{n_{N{a_2}C{O_3}}}}}$
[ ${{n_{N{a_2}C{O_3}}}}$ << ${{n_{{H_2}O}}}$ ]
${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}}}}$
$ \Rightarrow $ ${{24 - {P_S}} \over {24}} = {{3 \times 0.005} \over 2}$
$ \Rightarrow $ 24 – PS = 0.18
$ \Rightarrow $ PS = 23.82
Lowering in pressure ($\Delta $P) = 0.18 mm of Hg
= 18 × 10–2 mm of Hg
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ____________ molal. (Rounded off to the nearest integer)
[Kb = 0.52 K kg mol$-$1]
[Kb = 0.52 K kg mol$-$1]
Correct Answer: 3
Explanation:
As AB is a binary electrolyte,
$\therefore$ AB $\rightleftharpoons$ A+ + B$-$, n = 2
$i = 1 + \alpha (n - 1) = 1 + {{75} \over {100}}(2 - 1) = 1.75$
Given, $\Delta$Tb = 2.5 K
Kb = 0.52 K kg mol$-$1
$\therefore$ $\Delta$Tb = Kb $\times$ m $\times$ i
$ \Rightarrow m = {{\Delta {T_b}} \over {{K_b} \times i}} = {{2.5} \over {0.52 \times 1.75}}$
$ = 2.74 \simeq 3$ mol/kg
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
1 molal aqueous solution of an electrolyte A2B3 is 60% ionised. The boiling point of the solution at 1 atm is _________ K. (Rounded off to the nearest integer)
[Given Kb for (H2O) = 0.52 K kg mol$-$1]
[Given Kb for (H2O) = 0.52 K kg mol$-$1]
Correct Answer: 375
Explanation:
${A_2}{B_3}\buildrel {} \over
\longrightarrow 2{A^{ + 3}} + 3{B^{ - 2}}$
No. of ions = 2 + 3 = 5
i = 1 + (n $-$ 1) $\alpha $
= 1 + (5 $-$ 1) $\times$ 0.6
= 1 + 4 $\times$ 0.6 = 1 + 2.4 = 3.4
$\Delta {T_b} = {K_b} \times m \times i = 0.52 \times 1 \times 3.4 = 1.768^\circ $ C
$\Delta {T_b} = {({T_b})_{solution}} - {[{({T_b})_{{H_2}O}}]_{Solution}}$
$1.768 = {({T_b})_{solution}} - 100$
${({T_b})_{solution}} = 101.768^\circ $ C
$ = 375$ K
No. of ions = 2 + 3 = 5
i = 1 + (n $-$ 1) $\alpha $
= 1 + (5 $-$ 1) $\times$ 0.6
= 1 + 4 $\times$ 0.6 = 1 + 2.4 = 3.4
$\Delta {T_b} = {K_b} \times m \times i = 0.52 \times 1 \times 3.4 = 1.768^\circ $ C
$\Delta {T_b} = {({T_b})_{solution}} - {[{({T_b})_{{H_2}O}}]_{Solution}}$
$1.768 = {({T_b})_{solution}} - 100$
${({T_b})_{solution}} = 101.768^\circ $ C
$ = 375$ K
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
C6H6 freezes at 5.5$^\circ$C. The temperature at which a solution of 10g of C4H10 in 200g of C6H6 freeze is __________ $^\circ$C. (The molal freezing point depression constant of C6H6 is 5.12$^\circ$C/m.)
Correct Answer: 1
Explanation:
Pure solvent : C6H6(l)
Given,
$T_f^o = 5.5^\circ C$
${K_f} = 5.12^\circ C/m \Rightarrow m = 200g$
${m_{solute}} = 10g$
Molar mass of solute ${C_4}{H_{10}} = 12 \times 4 + 10 = 58$
Solute (C4H10) is non-dissociative;
$\therefore$ i = 1
$\therefore$ $\Delta {T_f} = i{K_f}\,m$
$ \Rightarrow (T_f^o - T_f^1) = 1 \times 5.12 \times {{(10/58)} \over {(200/1000)}}$
$5.5 - T_f^1 = {{5.12 \times 5 \times 10} \over {58}} \Rightarrow T_f^1 = 1.086^\circ C$
or $T_f^1 \approx 1^\circ C$
Given,
$T_f^o = 5.5^\circ C$
${K_f} = 5.12^\circ C/m \Rightarrow m = 200g$
${m_{solute}} = 10g$
Molar mass of solute ${C_4}{H_{10}} = 12 \times 4 + 10 = 58$
Solute (C4H10) is non-dissociative;
$\therefore$ i = 1
$\therefore$ $\Delta {T_f} = i{K_f}\,m$
$ \Rightarrow (T_f^o - T_f^1) = 1 \times 5.12 \times {{(10/58)} \over {(200/1000)}}$
$5.5 - T_f^1 = {{5.12 \times 5 \times 10} \over {58}} \Rightarrow T_f^1 = 1.086^\circ C$
or $T_f^1 \approx 1^\circ C$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
When 9.45 g of CICH2COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of CICH2COOH is x $ \times $ 10-3.
The value of x is ________. (Rounded off to the nearest integer)
[Kf(H20) = 1.86 K kg mol-1]
The value of x is ________. (Rounded off to the nearest integer)
[Kf(H20) = 1.86 K kg mol-1]
Correct Answer: 34.4
Explanation:
Here, $i = {{Final\,moles} \over {Initial\,moles}}$
i = 1 $-$ $\alpha$ + $\alpha$ + $\alpha$ $\Rightarrow$ i = 1 + $\alpha$
Formula used for freezing point; $\Delta$Tf = i Kfm
Here, Kf = freezing constant of H2O
Kf(H2O) = 1.86 K kg mol$-$1
m = molarity
$\Delta$Tf = i Kfm
0.5 = (1 + $\alpha$) (1.86)${{(9.45/94.5)} \over {(500/1000)}}$
${5 \over {3.72}}$ = 1 + $\alpha$ $\Rightarrow$ ${5 \over {3.72}}$ $-$ 1 = $\alpha$
$\Rightarrow$ $\alpha$ $ = {{1.28} \over {3.72}} = {{32} \over {93}} = 0.344$
Percentage of dissociation = 34.4%
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 1 Online
The value of x is ________.
Correct Answer: 100.1
Explanation:
$
\begin{aligned}
& \mathrm{AgNO}_3(\mathrm{aq}) \longrightarrow \underset{0.1 ~ m}{\mathrm{Ag}^{+}(\mathrm{aq})}+\underset{0.1 ~m}{\mathrm{NO}_3^{-}(\mathrm{aq})}\\\\
& \Delta \mathrm{T}_{\mathrm{b}}=0.2 \times 0.5 \\\\
& =0.1^{\circ} \mathrm{C}=0.1 \mathrm{~K}
\end{aligned}
$
Boiling point of solution $=100.1^{\circ} \mathrm{C} =\mathrm{X} $
Boiling point of solution $=100.1^{\circ} \mathrm{C} =\mathrm{X} $
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 1 Online
The value of | y | is ________.
Correct Answer: 2.5
Explanation:
$
\begin{aligned}
& \mathrm{AgNO}_3(\mathrm{aq}) \longrightarrow \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{NO}_3^{-}(\mathrm{aq}) \\\\
& 0.05 \mathrm{~m} \quad\quad\quad\quad\quad 0.05 \mathrm{~m} \quad\quad 0.05 \mathrm{~m} \\\\
& \mathrm{BaCl}_2(\mathrm{aq}) \longrightarrow \mathrm{Ba}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \\\\
& 0.05 \mathrm{~m} \quad \quad\quad\quad\quad0.05 \mathrm{~m} \quad\quad 0.1 \mathrm{~m}
\end{aligned}
$
$\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$combine to form $\mathrm{AgCl}$ precipitate
$ \begin{array}{cccc} & \mathrm{Ag}^{+}(\mathrm{aq})+ & \mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow & \mathrm{AgCl}(\mathrm{s}) \\\\ \mathrm{t}=0 & 0.05 \mathrm{~m} & 0.1 \mathrm{~m} \\\\ \mathrm{t}=\infty & 0 & 0.05 \mathrm{~m} \end{array} $
In final solution total concentration of all ions :
$ \begin{aligned} & {\left[\mathrm{Cl}^{-}\right]+\left[\mathrm{NO}_3^{-}\right]+\left[\mathrm{Ba}^{2+}\right]=0.05+0.05+0.05} \\\\ & \begin{aligned} \Delta \mathrm{T}_{\mathrm{b}} & =0.5 \times 0.15 \\\\ & =0.15 \mathrm{~m} \\\\ & =0.075^{\circ} \mathrm{C} \end{aligned} \end{aligned} $
B.P. of solution ' $\mathrm{B}$ ' $=100.075^{\circ} \mathrm{C}$
B.P. of solution ' $\mathrm{A}^{\prime}=100.1^{\circ} \mathrm{C}$
$|y|=100.1-100.075$
$ =0.025=2.5 \times 10^{-2} $
$\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$combine to form $\mathrm{AgCl}$ precipitate
$ \begin{array}{cccc} & \mathrm{Ag}^{+}(\mathrm{aq})+ & \mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow & \mathrm{AgCl}(\mathrm{s}) \\\\ \mathrm{t}=0 & 0.05 \mathrm{~m} & 0.1 \mathrm{~m} \\\\ \mathrm{t}=\infty & 0 & 0.05 \mathrm{~m} \end{array} $
In final solution total concentration of all ions :
$ \begin{aligned} & {\left[\mathrm{Cl}^{-}\right]+\left[\mathrm{NO}_3^{-}\right]+\left[\mathrm{Ba}^{2+}\right]=0.05+0.05+0.05} \\\\ & \begin{aligned} \Delta \mathrm{T}_{\mathrm{b}} & =0.5 \times 0.15 \\\\ & =0.15 \mathrm{~m} \\\\ & =0.075^{\circ} \mathrm{C} \end{aligned} \end{aligned} $
B.P. of solution ' $\mathrm{B}$ ' $=100.075^{\circ} \mathrm{C}$
B.P. of solution ' $\mathrm{A}^{\prime}=100.1^{\circ} \mathrm{C}$
$|y|=100.1-100.075$
$ =0.025=2.5 \times 10^{-2} $
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
A set of solutions is prepared using 180 g of
water as a solvent and 10 g of different nonvolatile solutes A, B and C. The relative
lowering of vapour pressure in the presence of
these solutes are in the order :
[Given, molar mass of A = 100 g mol–1; B = 200 g mol–1; C = 10,000 g mol–1]
[Given, molar mass of A = 100 g mol–1; B = 200 g mol–1; C = 10,000 g mol–1]
A.
A > C > B
B.
C > B > A
C.
A > B > C
D.
B > C > A
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
The size of a raw mango shrinks to a much
smaller size when kept in a concentrated salt
solution. Which one of the following processes
can explain this?
A.
Osmosis
B.
Reverse osmosis
C.
Diffusion
D.
Dialysis
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
A graph of vapour pressure and temperature for three different liquids X, Y, and Z is shown below :
The following inferences are made :
(A) X has higher intermolecular interactions compared to Y.
(B) X has lower intermolecular interactions compared to Y.
(C) Z has lower intermolecular interactions compared to Y.
The correct inference (s) is / are :
The following inferences are made :
(A) X has higher intermolecular interactions compared to Y.
(B) X has lower intermolecular interactions compared to Y.
(C) Z has lower intermolecular interactions compared to Y.
The correct inference (s) is / are :
A.
(B)
B.
(A)
C.
(C)
D.
(A) and (C)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
Two open beakers one containing a solvent and the other containing a mixture of that solvent
with a non voltatile solute are together selated in a container. Over time :
A.
The volume of the solution increases and the volume of the solvent decreases.
B.
The volume of solution does not change and the volume of the solvent decreases.
C.
The volumer of the solution and the solvent does not change.
D.
The volume of the solution decreases and the volume of the solvent increases.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
At 35oC, the vapour pressure of CS2
is 512 mm. Hg and that of acetone is 344 mm Hg. A solution
of CS2
in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the
following is :
A.
CS2 and acetone are less attracted to each other than to themselves.
B.
a mixture of 100 ml. CS2 and 100 ml. acetone has a volume < 200 ml.
C.
Heat must be absorved in order to produce the solution 35oC
D.
Raoult's law is not obeyed by this system.
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Morning Slot
The elevation of boiling point of 0.10 m
aqueous CrCl3.xNH3 solution is two times that
of 0.05 m aqueous CaCl2 solution. The value of
x is ______.
[Assume 100% ionisation of the complex and CaCl2, coordination number of Cr as 6, and that all NH3 molecules are present inside the coordination sphere]
[Assume 100% ionisation of the complex and CaCl2, coordination number of Cr as 6, and that all NH3 molecules are present inside the coordination sphere]
Correct Answer: 5
Explanation:
Molality of CaCl2 solution = 0.05 m
$\Delta $Tb = i Kb m = 3 × Kb × 0.05 = 0.15 Kb
Molality of CrCl3.xNH3 = 0.10 m
$\Delta $Tb' = i Kb $ \times $ 0.10
Given, $\Delta $Tb' = 2$\Delta $Tb
$ \Rightarrow $ i Kb $ \times $ 0.10 = 0.15 Kb
$ \Rightarrow $ i = 3
Co-ordination number of Cr is 6.
[Cr(NH3 )x.Cl6-x]Cl3-6+x $ \to $ [Cr(NH3 )xCl] + (3-6+x)Cl–
i = 3 = 1 + 3 - 6 + x
$ \Rightarrow $ x = 5
$\Delta $Tb = i Kb m = 3 × Kb × 0.05 = 0.15 Kb
Molality of CrCl3.xNH3 = 0.10 m
$\Delta $Tb' = i Kb $ \times $ 0.10
Given, $\Delta $Tb' = 2$\Delta $Tb
$ \Rightarrow $ i Kb $ \times $ 0.10 = 0.15 Kb
$ \Rightarrow $ i = 3
Co-ordination number of Cr is 6.
[Cr(NH3 )x.Cl6-x]Cl3-6+x $ \to $ [Cr(NH3 )xCl] + (3-6+x)Cl–
i = 3 = 1 + 3 - 6 + x
$ \Rightarrow $ x = 5
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Evening Slot
The osmotic pressure of a solution of NaCl is
0.10 atm and that of a glucose solution is
0.20 atm. The osmotic pressure of a solution
formed by mixing 1 L of the sodium chloride
solution with 2 L of
the glucose solution is x $ \times $ 10–3 atm. x is _____. (nearest integer)
the glucose solution is x $ \times $ 10–3 atm. x is _____. (nearest integer)
Correct Answer: 167
Explanation:
For NaCl : $\pi $1 = 0.1 atm, v1 = 1 L
For Glucose: $\pi $2 = 0.2 atm, v2 = 2 L
$\pi $mix = ${{{\pi _1}{v_1} + {\pi _2}{v_2}} \over {{v_1} + {v_2}}}$
= ${{0.1 \times 1 + 0.2 \times 2} \over {1 + 2}}$
= ${{0.5} \over 3}$ = ${{500} \over 3} \times {10^{ - 3}}$
= 167 $ \times $ 10-3 atm
For Glucose: $\pi $2 = 0.2 atm, v2 = 2 L
$\pi $mix = ${{{\pi _1}{v_1} + {\pi _2}{v_2}} \over {{v_1} + {v_2}}}$
= ${{0.1 \times 1 + 0.2 \times 2} \over {1 + 2}}$
= ${{0.5} \over 3}$ = ${{500} \over 3} \times {10^{ - 3}}$
= 167 $ \times $ 10-3 atm
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Morning Slot
At 300 K, the vapour pressure of a solution
containing 1 mole of n-hexane and 3 moles of
n-heptane is 550 mm of Hg. At the same
temperature, if one more mole of n-heptane is
added to this solution, the vapour pressure of
the solution increases by 10 mm of Hg. What is
the vapour pressure in mm Hg of n-heptane in
its pure state ______?
Correct Answer: 600
Explanation:
Let X1 and P$P_1^o$
are the mole fraction and vapour
pressure of n-hexane in solution and X2 and $P_2^o$
are the mole fraction and vapour pressure of
n-heptane in solution then
550 = $P_1^o$$ \times $${1 \over 4}$ + $P_2^o$$ \times $${3 \over 4}$
$ \Rightarrow $ 2200 = $P_1^o$ + 3$P_2^o$ ....(1)
On addition of 1 more mole of n-heptane
560 = $P_1^o$$ \times $${1 \over 5}$ + $P_2^o$$ \times $${4 \over 5}$
$ \Rightarrow $ 2800 = $P_1^o$ + 4$P_2^o$ ....(2)
From (1) and (2),
$P_1^o$ = 400, $P_2^o$ = 600
550 = $P_1^o$$ \times $${1 \over 4}$ + $P_2^o$$ \times $${3 \over 4}$
$ \Rightarrow $ 2200 = $P_1^o$ + 3$P_2^o$ ....(1)
On addition of 1 more mole of n-heptane
560 = $P_1^o$$ \times $${1 \over 5}$ + $P_2^o$$ \times $${4 \over 5}$
$ \Rightarrow $ 2800 = $P_1^o$ + 4$P_2^o$ ....(2)
From (1) and (2),
$P_1^o$ = 400, $P_2^o$ = 600
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Evening Slot
If 250 cm3
of an aqueous solution containing 0.73 g of a protein A is isotonic with one litre of
another aqueous solution containing 1.65 g of a protein B, at 298 K, the ratio of the molecular
masses of A and B is ______ × 10–2 (to the nearest integer).
Correct Answer: 177
Explanation:
Let molar mass of protein A = x g/mol
Let molar mass of protein B = y g/mol
$\pi $A = osmotic pressure of protein A = ${{0.73} \over x} \times {{1000} \over {250}} \times RT$
$\pi $B = osmotic pressure of protein B = ${{1.65} \over y} \times {1 \over 1} \times RT$
Given $\pi $A = $\pi $B
$ \Rightarrow $ ${{0.73} \over x} \times {{1000} \over {250}} \times RT$ = ${{1.65} \over y} \times {1 \over 1} \times RT$
$ \Rightarrow $ ${x \over y}$ = ${{0.73} \over {0.25 \times 1.65}}$ = 1.77
= 177 $ \times $ 10-2
Let molar mass of protein B = y g/mol
$\pi $A = osmotic pressure of protein A = ${{0.73} \over x} \times {{1000} \over {250}} \times RT$
$\pi $B = osmotic pressure of protein B = ${{1.65} \over y} \times {1 \over 1} \times RT$
Given $\pi $A = $\pi $B
$ \Rightarrow $ ${{0.73} \over x} \times {{1000} \over {250}} \times RT$ = ${{1.65} \over y} \times {1 \over 1} \times RT$
$ \Rightarrow $ ${x \over y}$ = ${{0.73} \over {0.25 \times 1.65}}$ = 1.77
= 177 $ \times $ 10-2
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Evening Slot
A cylinder containing an ideal gas (0.1 mol of
1.0 dm3) is in thermal equilibrium with a large
volume of 0.5 molal aqueous solution of
ethylene glycol at its freezing point. If the
stoppers S1 and S2 (as shown in the figure) are
suddenly withdrawn, the volume of the gas in
litres after equilibrium is achieved will be____.
(Given, Kf (water) = 2.0 K kg mol–1,
R = 0.08 dm3 atm K–1 mol–1)
(Given, Kf (water) = 2.0 K kg mol–1,
R = 0.08 dm3 atm K–1 mol–1)
Correct Answer: 2.17TO2.23
Explanation:
For aqueous solution
$\Delta $Tf = Kf.m = 2 × 0.5
$ \therefore $ Temperature of solution = –1°C = 272 K
$ \therefore $ Final volume of ideal gas = ${{nRT} \over P}$
= ${{0.1 \times 0.08 \times 272} \over 1}$
= 2.18 L
$\Delta $Tf = Kf.m = 2 × 0.5
$ \therefore $ Temperature of solution = –1°C = 272 K
$ \therefore $ Final volume of ideal gas = ${{nRT} \over P}$
= ${{0.1 \times 0.08 \times 272} \over 1}$
= 2.18 L
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Morning Slot
How much amount of NaCl should be added
to 600 g of water ($\rho $ = 1.00 g/mL) to decrease
the freezing point of water to – 0.2 °C ?
______.
(The freezing point depression constant for water = 2K kg mol–1)
(The freezing point depression constant for water = 2K kg mol–1)
Correct Answer: 1.74to1.76
Explanation:
$\Delta $Tf = 0.2o C
$ \therefore $ $\Delta $Tf = ikf.m
i = 2 for NaCl
$ \therefore $ 0.2 = 2$ \times $2$ \times $${{{W_{NaCl}} \times 1000} \over {58.5 \times 600}}$
$ \Rightarrow $ WNaCl = ${{58.5 \times 600 \times 0.2} \over {4 \times 1000}}$ = 1.76 g
$ \therefore $ $\Delta $Tf = ikf.m
i = 2 for NaCl
$ \therefore $ 0.2 = 2$ \times $2$ \times $${{{W_{NaCl}} \times 1000} \over {58.5 \times 600}}$
$ \Rightarrow $ WNaCl = ${{58.5 \times 600 \times 0.2} \over {4 \times 1000}}$ = 1.76 g
2020
JEE Mains
MSQ
JEE Main 2020 (Online) 3rd September Morning Slot
Henry’s constant (in kbar) for four gases $\alpha $, $\beta $, $\gamma $ and $\delta $ in water at 298 K is given below :
(density of water = 103 kg m-3 at 298 K)
This table implies that :
| $\alpha $ | $\beta $ | $\gamma $ | $\delta $ | |
|---|---|---|---|---|
| KH | 50 | 2 | 2 $ \times $ 10-5 | 0.5 |
(density of water = 103 kg m-3 at 298 K)
This table implies that :
A.
solubility of $\gamma $ at 308 K is lower than at 298 K
B.
The pressure of a 55.5 molal solution of $\delta $ is 250 bar
C.
$\alpha $ has the highest solubility in water at a given pressure
D.
The pressure of a 55.5 molal solution of $\gamma $ is 1 bar
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
Liquids A and B form ideal solution for all compositions of A and B at 25$^\circ $C. Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapour pressure of 0.3 and 0.4 bar, respectively. What is the vapour pressure of pure liquid B in bar?
Correct Answer: 0.2
Explanation:
Using Raoult's law equation for a mixture of volatile liquids.
${P_T} = p_A^o{\chi _A} + p_B^o{\chi _B}$
$0.3 = 0.25\chi p_A^o + 0.75\chi p_B^o$ ....(i)
$0.4 = 0.5\chi p_A^o + 0.5\chi p_B^o$ ....(ii)
By solving equation (i) and (ii)
$p_A^o$ = 0.6 bar and $p_B^o$ = 0.2 bar
Thus, the vapour pressure of pure liquid B in bar is 0.2.
${P_T} = p_A^o{\chi _A} + p_B^o{\chi _B}$
$0.3 = 0.25\chi p_A^o + 0.75\chi p_B^o$ ....(i)
$0.4 = 0.5\chi p_A^o + 0.5\chi p_B^o$ ....(ii)
By solving equation (i) and (ii)
$p_A^o$ = 0.6 bar and $p_B^o$ = 0.2 bar
Thus, the vapour pressure of pure liquid B in bar is 0.2.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1) and 1.8 g of glucose (molar
mass = 180 g mol–1) in 100 mL of water at 27oC. The osmotic pressure of the solution is :
(R = 0.08206 L atm K–1 mol–1)
(R = 0.08206 L atm K–1 mol–1)
A.
8.2 atm
B.
2.46 atm
C.
4.92 atm
D.
1.64 atm
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose
ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, ${{\Delta {T_b}(A)} \over {\Delta {T_b}(B)}}$, is :
A.
5 : 1
B.
1 : 0.2
C.
10 : 1
D.
1 : 5
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
At room temperature, a dilute solution of urea is prepared by dissolving 0.60 of urea in 360 g of water. If the
vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be.
(molar mass of urea = 60 g mol–1)
A.
0.031 mmHg
B.
0.017 mmHg
C.
0.028 mmHg
D.
0.027 mmHg
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
Molal depression constant for a solvent is
4.0 kg mol–1. The depression in the freezing
point of the solvent for 0.03 mol kg–1 solution
of K2SO4 is :
(Assume complete dissociation of the electrolyte)
(Assume complete dissociation of the electrolyte)
A.
0.18 K
B.
0.24 K
C.
0.36 K
D.
0.12 K
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
Liquid 'M' and liquid 'N' form an ideal solution.
The vapour pressures of pure liquids 'M' and
'N' are 450 and 700 mmHg, respectively, at the
same temperature. Then correct statement is:
(xM = Mole fraction of 'M' in solution ;
xN = Mole fraction of 'N' in solution ;
yM = Mole fraction of 'M' in vapour phase ;
yN = Mole fraction of 'N' in vapour phase)
(xM = Mole fraction of 'M' in solution ;
xN = Mole fraction of 'N' in solution ;
yM = Mole fraction of 'M' in vapour phase ;
yN = Mole fraction of 'N' in vapour phase)
A.
${{{x_M}} \over {{x_N}}} < {{{y_N}} \over {{y_N}}}$
B.
(xM – yM) < (xN – yN)
C.
${{{x_M}} \over {{x_N}}} = {{{y_N}} \over {{y_N}}}$
D.
${{{x_M}} \over {{x_N}}} > {{{y_M}} \over {{y_N}}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
The osmotic pressure of a dilute solution of an
ionic compound XY in water is four times that
of a solution of 0.01 M BaCl2 in water.
Assuming complete dissociation of the given
ionic compounds in water, the concentration of
XY (in mol L–1) in solution is :
A.
4 × 10–4
B.
6 × 10–2
C.
4 × 10–2
D.
16 × 10–4