1

### JEE Main 2016 (Online) 9th April Morning Slot

The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L−1. The solubility (in g L−1) at 750 torr partial pressure is :
A
0.0075
B
0.015
C
0.02
D
0.005

## Explanation

Partial pressure = Mole fraction × Solubility

${{{p_1}} \over {{p_2}}} = {{{s_1}} \over {{s_2}}}$

$\Rightarrow$ ${{500} \over {750}} = {{0.01} \over {{s_2}}}$

$\Rightarrow$ s2 = 0.015 g L-1
2

### JEE Main 2016 (Online) 10th April Morning Slot

An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is :
A
0.33
B
0.50
C
0.67
D
0.80
3

### JEE Main 2017 (Offline)

The freezing point of benzene decreases by 0.450C when 0.2 g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: (Kf for benzene = 5.12 K kg mol–1)
A
80.4 %
B
74.6 %
C
94.6 %
D
64.6 %

## Explanation

$\Delta$Tf = i $\times$ Kf $\times$ m

$\Rightarrow$ 0.45 = i $\times$ 5.12 $\times$ ${{0.2 \times 1000} \over {60 \times 20}}$

$\Rightarrow$ i = 0.527

2CH3COOH ⇌ (CH3COOH)2
1 - $\alpha$                   ${\alpha \over 2}$

i = 1 - $\alpha$ + ${\alpha \over 2}$

$\Rightarrow$ $\alpha$ = 0.946

$\therefore$ % dissociation is 94.6%.
4

### JEE Main 2017 (Online) 8th April Morning Slot

5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82oC. If Na2SO4 is 81.5% ionised, the value of x

(Kf for water=1.86oC kg mol−1) is approximately :

(molar mass of S = 32 g mol−1 and that of Na = 23 g mol−1)
A
15 g
B
25 g
C
45 g
D
65 g

## Explanation

Na2SO4 $\rightleftharpoons$ 2Na+ + SO4-2
Initial 1 mol 0 0
After dissociation 1 - x 2x x

$\therefore\,\,\,$ Total no of Moles after dissociation = 1 + 2x

Na2SO4 is ionised 81.5% means x = 0.815

Von't Hoff factor (i) = ${{Moles\,\,after\,\,dissociation} \over {Initial\,\,no.\,\,of\,\,moles}}$

= ${{1 + 2x} \over 1}$

= 1 + 2 $\times$ 0.815

= 2.63

$\therefore\,\,\,$ $\Delta$Tf = ${{1000 \times {K_f} \times {w_2} \times i} \over {{M_S} \times {w_1}}}$

$\Rightarrow $$\,\,\, 3.82 = {{1000 \times 1.86 \times 2.63 \times 5} \over {142 \times x}} \Rightarrow$$\,\,\,$ x = 45 gm