1
MCQ (Single Correct Answer)

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
2
MCQ (Single Correct Answer)

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%.
3
MCQ (Single Correct Answer)

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
4
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 9th April Morning Slot

A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl = 35.5 g mol−1)
A
0.162
B
0.675
C
0.325
D
0.486

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