1

AIPMT 2015 Cancelled Paper

The boiling point of 0.2 mol kg$-$1 solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?
A
Molecular mass of X is less than the molecular mass of Y.
B
Y is undergoing dissociation in water while X undergoes no change.
C
X is indergoing dissociation in water.
D
Molecular mass of X is greater than the molecular mass of Y.

Explanation

$\Delta$Tb = iKbm

Given, ($\Delta$Tb)x > ($\Delta$Tb)y

$\therefore$ ixKbm > iyKbm

$\Rightarrow$ ix > iy

(Kb is same for same solvent)

So, x is undergoing dissociation in water.
2

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. .
3

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.
4

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