Question

An unknown gas diffuses four times faster than oxygen . Calculate the molecular mass of the gas.

Answer 1

the answer to this is

Answer 2

"To solve the given problem, Graham's law related to rate of diffusion is needed. The “Graham's law" states that the "rate of diffusion" or "of effusion" of a "gas" is "inversely proportional" to the "square root" of its "molecular weight”. The "formula" can be written as:

$\frac { Rate\quad 1 }{ Rate\quad 2 } \quad =\quad \sqrt { \frac { { M }_{ 1 } }{ { M }_{ 2 } }}$

Where:

Rate 1 - "Rate of diffusion" of 1st gas.

Rate 2 - "Rate of diffusion" of 2nd gas.

${ M }_{ 1 }$ - Molar mass of 1st gas

${ M }_{ 2 }$ - Molar mass of 2nd gas

Given:  

Gas 1 - Unknown gas

Gas 2 - Oxygen gas

Molecular mass (${ M }_{ 2 }$) of oxygen is 32

As the gas 1 diffuses 4 times faster than oxygen, the  

$\frac { Rate\quad 1 }{ Rate\quad 2 } \quad =\quad 4$

Substituting the given value in Graham's equation:

$4\quad =\quad \sqrt { \frac { 32 }{ { M }_{ 1 } }}$

Squaring both sides we get:

$16\quad =\quad \frac { 32 }{ { M }_{ 1 } }$

${ M }_{ 1 }\quad =\quad \frac { 32 }{ 16 }$

${ M }_{ 1 }\quad =\quad 2$

In the "periodic table", the gas with molecular mass as 2 is hydrogen. Thus, it can be inferred that hydrogen diffuses "four times faster" than oxygen and the molecular mass of the gas is 2 gram/mol."