1

### AIPMT 2006

A solution of acetone in ethanol
A
obeys Raoult's law
B
shows a negative deviation from Raoult's law
C
shows a positive deviation from Raoult's law
D
behaves like a near ideal solution.

## Explanation

A solution of acetone in ethanol shows positive deviation from Raoult's law. It is because ethanol molecules are strongly hydrogen bonded. When acetone is added, these molecules break the hydrogen bonds and ethanol becomes more volatile. Therefore its vapour pressure is increased.
2

### AIPMT 2006

During osmosis, flow of water through a semipermeable membrane is
A
from solution having lower concentration only
B
from solution having higher concentration only
C
from both sides of semipermeable membrane with equal flow rates
D
from both sides of semipermeable membrane with unequal flow rates.

## Explanation

During osmosis, flow of water through semipermeable membrane is from solution having lower concentration only.
3

### AIPMT 2005

A solution of urea (mol. mass 56 g mol$-$1) boils at 100.18oC at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol$-$1 respectively, the above solution will freeze at
A
0.654oC
B
$-$ 0.654oC
C
6.54oC
D
$-$6.54oC

## Explanation

$\Delta$Tf = Kfm ......(1)

$\Delta$Tb = Kbm ......(2)

$\Rightarrow$ ${{\Delta {T_f}} \over {\Delta {T_b}}} = {{{K_f}} \over {{K_b}}}$ .....(3)

$\Delta$Tb = T2 – T1 = 100.18 – 100 = 0.18

kf for water = 1.86 K kg mol–1

kb for water = 0.512 K kg mol–1

${{\Delta {T_f}} \over {0.18}} = {{1.86} \over {0.512}}$

$\Rightarrow$ ${\Delta {T_f}}$ = 0.654

$\Delta$Tb = T2 – T1

$\Rightarrow$ 0.654 = 0 – T2

$\Rightarrow$ T2 = $-$ 0.654oC
4

### AIPMT 2005

A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20oC are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be
A
0.200
B
0.549
C
0.786
D
0.478

## Explanation

As the ratio of pentane to hexane = 1 : 4

$\therefore$ Mole fraction of pentane = 1/5

Mole fraction of hexane = 4/5

Total vapour pressure

= $\left( {{1 \over 5} \times 440 + {4 \over 5} \times 120} \right)$

= 184 mm of Hg

$\therefore$ Vapour pressure of pentane in mixture

= (Vapour pressure of mixture pentane $\times$ Mole fraction of in vapour phase)

$\Rightarrow$ 88 = 184 × mole fraction of pentane in vapour phase

$\Rightarrow$ Mole fraction of pentane in vapour phase

= ${{88} \over {184}}$ = 0.478