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During an diffusion experiment, it took 75 seconds for a certain number of moles of an unknown gas

to pass through a tiny hole. Under the same conditions, the same number of moles of oxygen gas

passed through the hole in 30 seconds. What is the molar mass of the unknown gas?

Verified Answer :

Answer:

200 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as volume of gas dispersed in a given interval of a time.

\(R=\frac{Volume}{time}\)

Number of moles of gas are directly proportional to the volume occupied by the gas.

\(moles \propto Volume\) (at constant temperature and pressure)

\(\frac{n_1}{V_1}=\frac{n_2}{V_2}\)

If n moles of unknown gas occupies V volume then n moles of oxygen will occupy:

\(V_1=\frac{n\times V}{n}=V\)

The n moles of unknown gas pass through a tiny hole in 75 seconds.The effusion rate is give as:

\(R=\frac{V}{75 s}\)

The n moles of oxygen gas pass through a tiny hole in 30 seconds.The effusion rate is give as:

\(R'=\frac{V}{30 s}\)

According to Graham law: the effusion rate is inversely proportional to the square root of molar mass of the gas.

\(\frac{R}{R'}=\sqrt{\frac{32 g/mol}{M}}\)

\(\frac{\frac{V}{75 s}}{\frac{V}{30 s}}=\sqrt{\frac{32 g/mol}{M}}\)

\(M=\frac{32 g/mol\times 75\times 75}{30 \times 30}=200 g/mol\)

200 g/mol is the molar mass of the unknown gas.

onyebuchinnaji

The molar mass of the unknown gas that diffused in the experiment is 200 g/mol.

The given parameters;

  • mass of the unknown gas = m₁
  • mass of oxygen gas, m₂ = 32 g/mol
  • time for oxygen gas diffusion, t₂ = 30 s
  • time for the unknown gas diffusion, t₁ = 75 s

The rate of diffusion of the gases is calculated using Graham's law of diffusion as follows;

\(\frac{R_1}{R_2} = \frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1} } \\\\(\frac{t_2}{t_1} )^2 = \frac{M_2}{M_1} \\\\(\frac{30}{75})^2 = \frac{32}{M_1} \\\\0.16 = \frac{32}{M_1} \\\\M_1 = \frac{32}{0.16} \\\\M_1 = 200 \ g/mol\)

Thus, the molar mass of the unknown gas that diffused in the experiment is 200 g/mol.

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