A volume of 100 ml of methane gas was confined in a diffusion tube and exposed to air, and at equilibrium a volume of 73.5 ml of air was measured in the tube. Again, when 100 ml of an unknown gas were

Question

A volume of 100 ml of methane gas was confined in a diffusion tube and exposed to air, and at equilibrium a volume of 73.5 ml of air was measured in the tube. Again, when 100 ml of an unknown gas were placed in the same tube and exposed to air, 104 ml of air were measured in the tube at equilibrium. The molecular mass of the unknown gas is

TestHub · Accepted Answer

According to Graham's Law of Effusion, the ratio of the effusion rates of two gases is inversely proportional to the square root of their molar masses. This can be expressed as:

$\frac{V _{ai r}}{V _{g a s}} = \frac{r _{ai r}}{r _{g a s}} = \frac{M _{g a s}}{M _{ai r}}$

By taking the ratio of two experiments, the molar mass of air ( $M_{ai r}$ ) can be eliminated. For an unknown gas (X) and methane (CH₄), the relationship is:

$\frac{M _{X}}{M _{C H_{4}}} = (\frac{V _{X} ( ai r )}{V _{C H_{4}} ( ai r )})^{2}$

Given the volumes, we can calculate the ratio:

$\frac{M _{X}}{M _{C H_{4}}} = (\frac{104}{73.5})^{2} \approx 2.00$

Since the molar mass of methane ( $M_{C H_{4}}$ ) is approximately 16 g/mol, the molar mass of the unknown gas ( $M_{X}$ ) is:

$M_{X} = 16 \times 2.00 = 32 g / m o l$

To verify this, we can first calculate $M_{ai r}$ using the methane data:

$M_{ai r} = 16 \times (\frac{100}{73.5})^{2} \approx 29.6 g / m o l$

Then, using $M_{ai r}$ and the data for gas X:

$M_{X} = 29.6 \times (\frac{104}{100})^{2} \approx 32.0 g / m o l$

Thus, the molecular mass of the unknown gas is 32 g/mol.

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