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1

AIPMT 2009

MCQ (Single Correct Answer)
A 0.0020 m aqueous solution of an ionic compound [Co(NH3)5(NO2)]Cl freezes at $$-$$ 0.00732oC. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = $$-$$1.86oC/m)
A
3
B
4
C
1
D
2

Explanation

The number of moles of ions produced by 1 mol of ionic compound = i

Given, m = 0.0020 m

$$\Delta $$Tf = 0oC – 0.00732oC = – 0.00732oC

Kf = – 1.86 oC

$$\Delta $$Tf = ikfm

$$ \Rightarrow $$ i = $${{0.00732} \over {1.86 \times 0.0020}}$$ = 2
2

AIPMT 2007

MCQ (Single Correct Answer)
0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol$$-$$1, the lowering in freezing point of the solution is
A
0.56 K
B
1.12 K
C
$$-$$ 0.56 K
D
$$-$$ 1.12 K

Explanation

HX H+ + X-
Initially 1 0 0
At equilibrium 1 - $$\alpha $$ $$\alpha $$ $$\alpha $$


Total moles = 1 – $$\alpha $$ + $$\alpha $$ + $$\alpha $$ = 1 + $$\alpha $$

$$ \therefore $$ i = $${{1 + \alpha } \over 1}$$

Given, $$\alpha $$ = 20% = 0.2

$$ \therefore $$ i = 1 + $$\alpha $$ = 1 + 0.2 = 1.2

$$\Delta $$Tf = ikf m = 1.2 × 1.86 × 0.5 = 1.12 K
3

AIPMT 2007

MCQ (Single Correct Answer)
Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g mL$$-$$1. Volume of acid required to make one litre of 0.1 M H2SO4 solution is
A
16.65 mL
B
22.20 mL
C
5.55 mL
D
11.10 mL

Explanation

Normality = $${{98 \times 1.8 \times 10} \over {49}}$$ = 36 N

N2 = 0.1 $$ \times $$ 2 = 0.2 N

N2V2 = N1V1

$$ \Rightarrow $$ 36 $$ \times $$ V = 0.2 $$ \times $$ 1000

$$ \Rightarrow $$ V = $${{0.2 \times 1000} \over {36}}$$ = 5.55 mL
4

AIPMT 2006

MCQ (Single Correct Answer)
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

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