1
MCQ (Single Correct Answer)

### NEET 2017

If molality of the dilute solution is doubled, the value of molal depression constant (Kf) will be
A
halved
B
tripled
C
unchanged
D
doubled.

## Explanation

The value of molal depression constant, Kf is constant for a particular solvent, thus, it will be unchanged when molality of the dilute solution is doubled.
2
MCQ (Single Correct Answer)

### NEET 2016 Phase 1

Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25oC. (Given, vapour pressure data at 25oC, benzene = 12.8 kPa, toluene = 3.85 kPa)
A
The vapour eill contain equal amounts of benzene and toluene.
B
Not enough information is given to make a prediction.
C
The vapour will contain a higher percentage of benzene.
D
The vapour will contain a higher percentage of toluene

## Explanation

pBenzene = xBenzene. poBenzene

pToluene = xToluene. poToluene

For an ideal 1 : 1 molar mixture of benzene and toluene

xBenzene = ${1 \over 2}$ and xToluene = ${1 \over 2}$

pBenzene = ${1 \over 2}$poBenzene = ${1 \over 2} \times 12.8$ = 6.4 kPa

pToluene = ${1 \over 2}$poToluene = ${1 \over 2} \times 3.85$ = 1.925 kPa

Thus, the vapour will contain a high percentage of benzene as the partial vapour pressure of benzene is higher as compared to that of toluene.
3
MCQ (Single Correct Answer)

### NEET 2016 Phase 1

At 100oC the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be
A
102oC
B
103oC
C
101oC
D
100oC

## Explanation

Given that

ws = 6.5 g, wA = 100 g

ps = 732 mm of Hg

kb = 0.52, Tob = 100oC

po = 760 mm of Hg

${{{p^o} - {p_s}} \over {{p^o}}} = {{{n_2}} \over {{n_1}}}$

$\Rightarrow$ ${{760 - 732} \over {760}} = {{{n_2}} \over {{{100} \over {18}}}}$

$\Rightarrow$ n2 = 0.2046 mol

$\Delta$Tb = Kb × m

Tb - Tob = ${k_b} \times {{{n_2} \times 1000} \over {{w_{A\left( g \right)}}}}$

$\Rightarrow$ Tb - 100oC = ${{0.52 \times 0.2046 \times 1000} \over {100}}$ = 1.06

$\Rightarrow$ Tb = 101.06 oC
4
MCQ (Single Correct Answer)

### NEET 2016 Phase 2

The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
A
0
B
1
C
2
D
3

## Explanation

Ba(OH)2 is a strong electrolyte and undergoes cent percent dissociation in a dilute aqueous solution.

Ba(OH)2(aq) $\to$ Ba2+(aq) + 2OH(aq)

Thus, van’t Hoff factor i = 3.

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