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1

### 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
2

### 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

### NEET 2016 Phase 2

Which one of the following is incorrect for ideal solution?
A
$$\Delta {H_{mix}} = 0$$
B
$$\Delta {U_{mix}} = 0$$
C
$$\Delta P = {P_{obs}} - P$$calculated by Raoult's law
D
$$\Delta {G_{mix}} = 0$$

## Explanation

For ideal solution, we have

$$\Delta$$Hmix = 0, $$\Delta$$Vmix = 0

Now Umix = ∆Hmix – P$$\Delta$$Vmix

$$\therefore$$ $$\Delta$$Umix = 0

Also, for an ideal solution,

pA = xApAo, pB = xBpBo

$$\therefore$$ $$\Delta$$p = pobserved – pcalculated = 0

$$\Delta$$Gmix = $$\Delta$$Hmix – T$$\Delta$$Smix

For an ideal solution, $$\Delta$$Smix $$\ne$$ 0

$$\therefore$$ $$\Delta$$Gmix $$\ne$$ 0
4

### 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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Class 12