Question

A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure effuses in 2.42 min. What is the molar mass of the unknown gas

Answer 1

Answer:

Molar mass of the gas is 0.0961 g/mol

Explanation:

The effusion rate of an unknown gas = 11.1 min

rate of $H_{2}$ effusion = 2.42 min

molar mass of hydrogen = 1 x 2 = 2 g/m

molar mas of unknown gas = ?

From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.

from

$\frac{R_{g} }{R_{h} }$ = $\sqrt{\frac{M_{h} }{M_{g} } }$

where

$R_{h}$ = rate of effusion of hydrogen gas

$R_{g}$ = rate of effusion of unknown gas

$M_{h}$ = molar mass of H2 gas

$M_{g}$ = molar mass of unknown gas

substituting values, we have

$\frac{11.1 }{2.42 }$ = $\sqrt{\frac{2 }{M_{g} } }$

4.587 = $\sqrt{\frac{2 }{M_{g} } }$

$\sqrt{M_{g} }$ = $\sqrt{2}$/4.587

$\sqrt{M_{g} }$ = 0.31

$M_{g}$ = $0.31^{2}$ = 0.0961 g/mol

Answer 2

The molar mass of the unknown gas will be "0.0961 g/mol".

Given:

Effusion rate of unknown gas,

• $R_g = 11.1 \ min$

Effusion rate of $H_2$,

• $R_h = 2.42 \ min$

Molar mass of hydrogen,

• $M_h = 1\times 2$

              $= 2 \ g/m$

According to the Graham's law, we get

→    $\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }$

By substituting the values, we get

→   $\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }$

→ $4.587=\sqrt{\frac{2}{M_g} }$

→ $\sqrt{M_g} = \sqrt{\frac{2}{4.587} }$

   $\sqrt{M_g} = 0.31$

       $M_g = 0.0961 \ g/mol$

Thus the above solution is right.          

Learn more:

https://brainly.com/question/6019799