Answer : The molecular weight of the unknown gas is, 57.0 amu
Explanation :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
[tex]R\propto \sqrt{\frac{1}{M}}[/tex]
And the relation between the rate of effusion and volume is :
[tex]R=\frac{V}{t}[/tex]
From this we conclude that,
[tex]\frac{V_1}{V_2}=\sqrt{\frac{M_2}{M_1}}[/tex]
where,
[tex]V_1[/tex] = rate of effusion of unknown gas = 11.50 mL
[tex]V_2[/tex] = rate of effusion of argon gas = 9.63 mL
[tex]M_1[/tex] = molar mass of unknown gas
[tex]M_2[/tex] = molar mass of argon gas = 40 amu
Now put all the given values in the above formula, we get:
[tex](\frac{11.50mL}{9.63mL})=\sqrt{\frac{M_1}{40amu}}[/tex]
[tex]M_1=57.0amu[/tex]
Therefore, the molecular weight of the unknown gas is, 57.0 amu
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Explanation:
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