Question

100 ml of O2 gas diffuses through a porous membrane in 20 seconds. 200 ml of gas ‘X’ takes 40 seconds to diffuse through same membrane under identical conditions. Molecular weight of gas ‘X’ is​

Answer 1

Answer:

1289

Explanation:

Rate & diffusion of gas O² = 100 ml

Rate & diffusion of gas X = 200 ml

mass & gass (M1) O²= 32 gm

mass & gass (M2) O²= ? gm

formula :

$rate1 \div rate2 = \sqrt{m1 \div m2}$

$100 \div 200 = \sqrt{32 \div m2}$

$1 \div 2 = 4 \sqrt2 \div \sqrt{m2}$

$\sqrt{m2} = 4 \sqrt{2} \times 2$

$\sqrt{m2} = 8 \sqrt{2}$

$m2 = 8 {}^{2} \times 2$

$m2 = 64 \times 2$

$m2 = 128$

X is Iodine = 128

Answer 2

Given:

100 ml of O2 gas diffuses through a porous membrane in 20 seconds. 200 ml of gas ‘X’ takes 40 seconds to diffuse through same membrane under identical conditions.

To find:

Molecular weight of X .

Calculation:

Let rate be R , time be t and molecular mass be m;

$R \: \propto \: \dfrac{1}{ \sqrt{m} } \: \propto \: \dfrac{1}{t}$

$\boxed{ = > t \: \propto \: \sqrt{m} }$

So, taking corresponding ratios:

$\sf{ \therefore \: \dfrac{t_{oxygen}}{t_{X}} = \dfrac{ \sqrt{ m_{oxygen}} }{ \sqrt{ m_{X}} } }$

$\sf{ = > \: \dfrac{20}{40} = \dfrac{ \sqrt{ m_{oxygen}} }{ \sqrt{ m_{X}} } }$

$\sf{ = > \: \dfrac{20}{40} = \dfrac{ \sqrt{32} }{ \sqrt{ m_{X}} } }$

$\sf{ = > \: \dfrac{1}{2} = \dfrac{ \sqrt{32} }{ \sqrt{ m_{X}} } }$

$\sf{ = > \: \dfrac{1}{4} = \dfrac{ 32}{ m_{X}} }$

$\sf{ = > \: m_{X} = 32 \times 4}$

$\sf{ = > \: m_{X} = 128 \: g {mol}^{ - 1} }$

So, final answer is:

$\boxed{ \rm{\: m_{X} = 128 \: g {mol}^{ - 1} }}$