1
MCQ (Single Correct Answer)

AIPMT 2006

1.00 g of a non-electrolyte solute (molar mass 250 g mol$$-$$1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol$$-$$1, the freezing point of benzene will be lowered by
A
0.2 K
B
0.4 K
C
0.3 K
D
0.5 K

Explanation

Molality of non-electrolyte solute

= $${1 \over {250 \times 0.0512}}$$ = 0.0781 m

$$\Delta $$Tf = kf m

= 5.12 × 0.0781 = 0.4 K
2
MCQ (Single Correct Answer)

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.
3
MCQ (Single Correct Answer)

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.
4
MCQ (Single Correct Answer)

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

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