1
MCQ (Single Correct Answer)

AIPMT 2014

Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?
A
KCl
B
C6H12O6
C
Al2(SO4)3
D
K2SO4

Explanation

We know that depression in freezing point ($$\Delta $$Tf ) is given as

$$\Delta $$Tf = iKfm

So, $$\Delta $$Tf $$ \propto $$ i

Thus, more value of i (Van’t Hoff factor), more will be depression in freezing point.

Al2(SO4)3 ⇌ 2Al+3 + 3SO42–

i is maximum i.e., 5 for Al2(SO4)3. .
2
MCQ (Single Correct Answer)

NEET 2013 (Karnataka)

Which condition is not satisfied by an ideal solution?
A
$${\Delta _{mix}}\,V = 0$$
B
$${\Delta _{mix}}\,S = 0$$
C
Obeyance to Raoult's Law
D
$${\Delta _{mix}}\,H = 0$$

Explanation

An ideal solution is follow:

1. Volume change ($$\Delta $$V) of mixing should be zero

2. Heat change ($$\Delta $$H) on mixing should be zero.

3. Obey Raoult’s law at every range of concentration.
3
MCQ (Single Correct Answer)

NEET 2013

How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO3? The concentrated acid is 70% HNO3.
A
70.0 g conc. HNO3
B
54.0 g conc. HNO3
C
45.0 g conc. HNO3
D
90.0 g conc. HNO3

Explanation

$$2 = {m \over {63 \times 0.25}}$$

$$ \Rightarrow $$ m = 2 × 63 × 0.25 = 31.5 g

Now, if concentrated HNO3 is 100% then it requires 31.5 g. But the original solution of HNO3 is 70% concentrated.

Hence, the mass of HNO3 required to produce 2.0 M solution

= $${{100} \over {75}} \times 31.5$$

= 45 g of HNO3
4
MCQ (Single Correct Answer)

AIPMT 2012 Mains

Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25oC are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same temperature will be
(Molecular mass of CHCl3 = 119.5 u and molecular mass of CH2Cl2 = 85 u)
A
173.9 mm Hg
B
615.0 mm Hg
C
90.6 mm Hg
D
285.5 mm Hg

Explanation

nCHCl3 = $${{25.5} \over {119.5}}$$ = 0.213

nCH2Cl2 = $${{40} \over {85}}$$ = 0.47

PT = PoAXA + PoBXB

= $$200 \times {{0.213} \over {0.683}} + 41.5 \times {{0.47} \over {0.683}}$$

= 62 + 28.55

= 90.63

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