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1

### AIPMT 2011 Prelims

The van't Hoff factor $$i$$ for a compound which undergoes dissociation in one solvent and association in other solvent is respectively
A
less than one and greater than one
B
less than one and less than one
C
greater than one and less than one
D
greater than one and greater than one.

## Explanation

From the value of van’t Hoff factor i it is possible to determine the degree of dissociation or association. In case of dissociation, i is greater than 1 and in case of association i is less than 1.
2

### AIPMT 2011 Mains

200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 $$\times$$ 10$$-$$3 bar. The molar mass of protein will be (R = 0.083 L bar mol$$-$$1 K$$-$$1)
A
51022 g mol$$-$$1
B
122044 g mol$$-$$1
C
31011 g mol$$-$$1
D
61038 g mol$$-$$1

## Explanation

Osmotic pressure, $$\pi$$ = CRT

$$\Rightarrow$$ $$\pi$$ = $${n \over V}$$RT

$$\Rightarrow$$ $$\pi$$V = $${w \over M}$$RT

M = $${{wRT} \over {\pi V}}$$

= $${{1.26 \times 0.083 \times 300} \over {2.57 \times {{10}^{ - 3}} \times {{200} \over {1000}}}}$$

= $${{1.26 \times 0.083 \times 300} \over {2.57 \times {{10}^{ - 3}} \times 0.2}}$$

= 61038 g mol–1
3

### AIPMT 2010 Prelims

An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?
A
B
C
D

## Explanation

Addition of solute decreases the vapour pressure as some sites of the surface are occupied by solute particles resulting in decreased surface area. However, addition of solvent, i.e., dilution increases the surface area of the liquid surface, thus results in increased vapour pressure. Hence, addition of water to the aqueous solution of (1 molal) KI results in increased vapour pressure.
4

### 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

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