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1

### AIPMT 2010 Prelims

A solution of sucrose (molar mass = 342 g mol$$-$$) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol$$-$$1)
A
$$-$$ 0.372oC
B
$$-$$ 0.520oC
C
+ 0.372oC
D
$$-$$ 0.570oC

## Explanation

Depression in freezing point,

$$\Delta$$Tf = Kf $$\times$$ m

m = $${{{w_B}} \over {{M_B}}} \times {{1000} \over {{W_A}}}$$

= $${{68.5 \times 1000} \over {342 \times 1000}}$$

$$\Delta$$Tf = 1.86 $$\times$$ $${{68.5} \over {342}}$$

= 0.372 oC

$$\therefore$$ Tf = 0 - 0.372 oC = - 0.372 oC
2

### AIPMT 2009

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
3

### AIPMT 2007

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
4

### AIPMT 2007

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

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