Respuesta :
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.
[tex]R=\frac{Volume}{time}[/tex]
Number of moles of gas are directly proportional to the volume occupied by the gas.
[tex]moles \propto Volume[/tex] (at constant temperature and pressure)
[tex]\frac{n_1}{V_1}=\frac{n_2}{V_2}[/tex]
If n moles of unknown gas occupies V volume then n moles of oxygen will occupy:
[tex]V_1=\frac{n\times V}{n}=V[/tex]
The n moles of unknown gas pass through a tiny hole in 75 seconds.The effusion rate is give as:
[tex]R=\frac{V}{75 s}[/tex]
The n moles of oxygen gas pass through a tiny hole in 30 seconds.The effusion rate is give as:
[tex]R'=\frac{V}{30 s}[/tex]
According to Graham law: the effusion rate is inversely proportional to the square root of molar mass of the gas.
[tex]\frac{R}{R'}=\sqrt{\frac{32 g/mol}{M}}[/tex]
[tex]\frac{\frac{V}{75 s}}{\frac{V}{30 s}}=\sqrt{\frac{32 g/mol}{M}}[/tex]
[tex]M=\frac{32 g/mol\times 75\times 75}{30 \times 30}=200 g/mol[/tex]
200 g/mol is the molar mass of the unknown gas.
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;
[tex]\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[/tex]
Thus, the molar mass of the unknown gas that diffused in the experiment is 200 g/mol.
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