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1

### AIPMT 2005

A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20oC are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be
A
0.200
B
0.549
C
0.786
D
0.478

## Explanation

As the ratio of pentane to hexane = 1 : 4

$$\therefore$$ Mole fraction of pentane = 1/5

Mole fraction of hexane = 4/5

Total vapour pressure

= $$\left( {{1 \over 5} \times 440 + {4 \over 5} \times 120} \right)$$

= 184 mm of Hg

$$\therefore$$ Vapour pressure of pentane in mixture

= (Vapour pressure of mixture pentane $$\times$$ Mole fraction of in vapour phase)

$$\Rightarrow$$ 88 = 184 × mole fraction of pentane in vapour phase

$$\Rightarrow$$ Mole fraction of pentane in vapour phase

= $${{88} \over {184}}$$ = 0.478
2

### AIPMT 2005

The mole fraction of the solute in one molal aqueous solution is
A
0.009
B
0.018
C
0.027
D
0.036

## Explanation

One molal solution means one mole solute present in 1 kg (1000 g) solvent.

$$\therefore$$ mole of solute = 1

Mole of solvent (H2O) = $${{1000} \over {18}}$$

Xsolute = $${1 \over {1 + {{1000} \over {18}}}}$$ = 0.018
3

### AIPMT 2005

The vapour pressure of two liquids P and Q are 80 and 60 torr, respectively. The total vapour pressure of solution obtained by mixing 3 mole of P and 2 mole of Q would be
A
72 torr
B
140 torr
C
68 torr
D
20 torr

## Explanation

Hence total vapour pressure

= [(Mole fraction of P) × (Vapour pressure of P)] + [(Mole fraction of Q) × Vapour pressure of Q)]

= $$\left( {{3 \over 5} \times 80 + {2 \over 5} \times 60} \right)$$

= 48 + 24 = 72 torr
4

### AIPMT 2002

2.5 litre of 1 M NaOH solution is mixed with another 3 litre of 0.5 M NaOH solution. Then find out molarity of resultant solution.
A
0.80 M
B
1.0 M
C
0.73 M
D
0.50 M

## Explanation

From molarity equation

M1V1 + M2V2 = MV

$$\Rightarrow$$ 1× 2.5 + 0.5 × 3 = M × 5.5

$$\Rightarrow$$ M = $${4 \over {5.5}}$$ = 0.73M

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