1

### AIPMT 2011 Prelims

In Duma's method of estimation of nitrogen 0.35 g of an organic compound gave 55 mLof nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be (aqueous tension at 300 K = 15 mm).
A
15.45
B
16.45
C
17.45
D
14.45

## Explanation

Given V1 = 55 mL, V2 = ?

P1 = 715 – 15 = 700 mm, P2 = 760 mm

T1 = 300 K, T2 = 273 K

General gas equation,

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$

$\Rightarrow$ ${{700 \times 55} \over {300}} = {{760 \times {V_2}} \over {273}}$

$\Rightarrow$ V2 = 46.098 mL

Now, 22400 mL of nitrogen = 1 mole

$\therefore$ 46.098 mL of nitrogen = ${{1 \times 46.098} \over {22400}}$ mol

Weight of nitrogen = ${{1 \times 46.098} \over {22400}}$ $\times$ 28 = 0.057 g

Percent composition of nitrogen in 0.35 g of compound

= ${{0.0576} \over {0.35}} \times 100$ = 16.45 %
2

### AIPMT 2011 Prelims

Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will be
A
50.00 u
B
12.25 u
C
6.50 u
D
25.00 u

## Explanation

We know that ${{{r_A}} \over {{r_B}}} = {{v/{t_A}} \over {v/{t_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}}$

$\Rightarrow {{{t_A}} \over {{t_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}} \Rightarrow {{10} \over {20}} = \sqrt {{{{M_B}} \over {49}}}$

$\Rightarrow {\left( {{{10} \over {20}}} \right)^2} = {{{M_B}} \over {49}} \Rightarrow {{100} \over {400}} = \,{{{M_B}} \over {49}}$

$\Rightarrow {M_B} = {{49 \times 100} \over {400}} = 12.25$ u
3

### AIPMT 2011 Mains

A bubble of air is underwater at temperature 15oC and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25oC and the pressure is 1.0 bar, what will happen to the volume of the bubble?
A
Volume will become greater by a factor of 1.6.
B
Volume will become greater by a factor of 1.1
C
Volume will become smaller by a factor of 0.70
D
Volume will become greater by a factor of 2.5

## Explanation

Given

P1 = 1.5 bar, T1 = 273 + 15 = 288 K, V1 = V

P2 = 1.0 bar, T1 = 273 + 25 = 298K, V2 = ?

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$

$\Rightarrow$ ${{1.5 \times V} \over {288}} = {{1 \times {V_2}} \over {298}}$

$\Rightarrow$ V2 = 1.55V

$\therefore$ Volume of bubble will be almost 1.6 time to initial volume of bubble.
4

### AIPMT 2011 Prelims

A gaseous mixture was prepared by taking equal mole of CO and N2. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N2) in the mixture is
A
0.5 atm
B
0.8 atm
C
0.9 atm
D
1 atm

## Explanation

Number of moles $\Rightarrow$ nCO = nN2

Volume of container is same. So, VCO = VN2 .

Also, temperature is same for both the gases thus, TCO = TN2

According to ideal gas equation, PV = nRT

Now, V, n, R and T for both gases are same. So, PCO = PN2

Now, total pressure is 1 atm and according to Dalton’s sum of partial pressure,

PCO + PN2 = 1 atm

$\Rightarrow$ 2PN2 = 1 atm {$\because$ PCO = PN2 }

$\Rightarrow$ PN2 = 0.5 atm