1
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
2
MCQ (Single Correct Answer)

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
3
MCQ (Single Correct Answer)

AIPMT 2005

The vapour pressure of two liquids P and Q are 80 and 60 torr, respectively. The total vapour pressure of solution obtained by mixing 3 mole of P and 2 mole of Q would be
A
72 torr
B
140 torr
C
68 torr
D
20 torr

Explanation

Hence total vapour pressure

= [(Mole fraction of P) × (Vapour pressure of P)] + [(Mole fraction of Q) × Vapour pressure of Q)]

= $$\left( {{3 \over 5} \times 80 + {2 \over 5} \times 60} \right)$$

= 48 + 24 = 72 torr
4
MCQ (Single Correct Answer)

AIPMT 2005

The mole fraction of the solute in one molal aqueous solution is
A
0.009
B
0.018
C
0.027
D
0.036

Explanation

One molal solution means one mole solute present in 1 kg (1000 g) solvent.

$$ \therefore $$ mole of solute = 1

Mole of solvent (H2O) = $${{1000} \over {18}}$$

Xsolute = $${1 \over {1 + {{1000} \over {18}}}}$$ = 0.018

EXAM MAP

Joint Entrance Examination

JEE Advanced JEE Main

Medical

NEET

Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI