1

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.
2

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
3

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.
4

AIPMT 2015 Cancelled Paper

Which one of the following electrolytes has the same value of van't Hoff factor (i) as that of Al2(SO4)3 (if all are 100% ionised)?
A
Al(NO3)3
B
K4[Fe(CN)6]
C
K2SO4
D
K3[Fe(CN)6]

Explanation

Ai2(SO4)3 ⇌ 2Al+3 + 3SO42–

van't Hoff factor, i = 5

K2SO4 ⇌ 2K+ + SO42–

van't Hoff factor, i = 3

K3[Fe(CN)6] ⇌ 3K+ + [Fe(CN)6]2-

van't Hoff factor, i = 4

Al(NO3)3 ⇌ Al3+ + 3NO3

van't Hoff factor, i = 4

K4[Fe(CN)6] ⇌ 4K+ + [Fe(CN)6]4–

van't Hoff factor, i = 5