1
MCQ (Single Correct Answer)

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
2
MCQ (Single Correct Answer)

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
3
MCQ (Single Correct Answer)

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

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

$$ \Rightarrow $$ $$\pi V = nRT$$

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

$$ \Rightarrow $$ $${M_2} = \left( {{{{m_2}} \over V}} \right){{RT} \over \pi }$$
4
MCQ (Single Correct Answer)

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.

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