1

### 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
2

### 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
3

### AIPMT 2012 Prelims

pA and pB are the vapour pressure of pure liquid components, A and B, respectively of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be
A
${p_A} + {x_A}\left( {{p_B} - {p_A}} \right)$
B
${p_A} + {x_A}\left( {{p_A} - {p_B}} \right)$
C
${p_B} + {x_A}\left( {{p_B} - {p_A}} \right)$
D
${p_B} + {x_A}\left( {{p_A} - {p_B}} \right)$

## Explanation

p = pAxA + pBxB

= pAxA + pB(1 – xA)

(As for binary solution xA + xB = 1)

= pAxA + pB – pBxA

= pB + xA(pA – pB)
4

### AIPMT 2011 Mains

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86oC/m, the freezing point of the solution will be
A
$-$0.18oC
B
$-$0.54oC
C
$-$0.36oC
D
$-$0.24oC

## Explanation

We know that $\Delta$Tf = i × Kf × m

Here i is van’t Hoff’s factor.

i for weak acid is 1 + $\alpha$.

Here $\alpha$ is degree of dissociation i.e., 30/100 = 0.3

$\therefore$ i = 1 + $\alpha$ = 1 + 0.3 = 1.3

$\therefore$ Tf = i × Kf × m = 1.3 × 1.86 × 0.1 = 0.24

$\therefore$ Freezing point = – 0.24