1

### AIPMT 2012 Prelims

50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is 36, the molecular mass of gass A will be
A
96
B
128.74
C
20.25
D
64.42

## Explanation

According to Graham's law of diffusion,

${{{r_1}} \over {{r_2}}} = \sqrt {{{{d_2}} \over {{d_1}}}} = \sqrt {{{{M_2}} \over {{M_1}}}}$

${r_A} = {{{V_A}} \over {{T_A}}},\,{r_B} = {{{V_B}} \over {{T_B}}}$

${{{V_A}/{T_A}} \over {{V_B}/{T_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}}$

${V_A} = {V_B},\,{T_A} = 150\,\sec$, TB = 200 sec, MB = 36, MA = ?

${{{T_B}} \over {{T_A}}} = \sqrt {{{{M_B}} \over {{M_A}}}} \Rightarrow {{200} \over {150}} = \sqrt {{{36} \over {{M_A}}}}$

$\Rightarrow {4 \over 3} = \sqrt {{{36} \over {{M_A}}}} \,\,or\,\,{{4 \times 4} \over {3 \times 3}} = {{36} \over {{M_A}}}$

$\Rightarrow$ MA = ${{36} \over {4 \times 4}} \times 3 \times 3 = 20.25$
2

### 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 %
3

### 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
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