Question

The rate of diffusion of an unknown gas is 1.73 times smaller than rate of diffusion of oxygen. Find molar mass of unknown gas.

Answer 1

from Graham's law of diffusion of gas,

rate of diffusion ∝ 1/√(molecular weight)

i.e., $\bf{\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}}$

a/c to question, rate of diffusion of unknown gas , X is 1.73 times smaller than rate of diffusion of oxygen.

i.e., $r_X=\frac{r_{O_2}}{1.73}$

or, $r_1=\frac{r_2}{1.73}$

or, $\frac{r_1}{r_2}=\frac{1}{1.73}$

molecular weight of oxygen, $M_2$ = 32g/mol

now, $\frac{1}{1.73}=\sqrt{\frac{32}{M_1}}$

squaring both sides,

or, 1/3 = 32/$M_1$

or, $M_1$ = 32 × 3 = 96 g/mol.

hence, molecular mass of unknown gas is 96 g/mol.

Answer 2

Answer:

96g

Explanation:

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