1

### 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
2

### AIPMT 2002

A solution contains non volatile solute of molecular mass M2. Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure?
A
${M_2} = \left( {{{{m_2}} \over \pi }} \right)VRT$
B
${M_2} = \left( {{{{m_2}} \over V}} \right){{RT} \over \pi }$
C
${M_2} = \left( {{{{m_2}} \over V}} \right)\pi RT$
D
${M_2} = \left( {{{{m_2}} \over V}} \right){\pi \over {RT}}$

## Explanation

For dilute solution, the van’t Hoff equation is

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

$\Rightarrow$ $\pi V = nRT$

$\Rightarrow$ $\pi V = {{{m_2}} \over M}RT$

$\Rightarrow$ ${M_2} = \left( {{{{m_2}} \over V}} \right){{RT} \over \pi }$
3

### AIPMT 2002

A solution containing components A and B folloes Raoult's law
A
A - B attraction force is greater than A - A and B - B
B
A - B attraction force is less than A - A and B - B
C
A - B attraction force remains same as A - A and B - B
D
volume of solution is different from sum of volume of solute and solvent.

## Explanation

These two components A and B follows the condition of Raoult’s law if the force of attraction between A and B is equal to the force of attraction between A and A or B and B.
4

### AIPMT 2001

Pure water can be obtained from sea water by
A
centrifugation
B
plasmolysis
C
reverse osmosis
D
sedimentation

## Explanation

The osmotic pressure of sea water is 25 atm at 15°C. When pressure greater than 26 atm is applied on sea water separated by a rigid semipermeable membrane. Pure water is obtained. This is also called desalination of sea water.