Question

a sample of he gas is observed to effuse through a porous barrier in 5.06 minutes. under the same conditions, the same number of moles of an unknown gas requires 17.5 minutes to effuse through the same barrier. the molar mass of the unknown gas is g/mol.
use the references to access important values if needed for this question.
submit answer
try another version
3 item attempts remaining

Explanation:

Step1: Calculate the rates of effusion

The rate of effusion \(r\) is inversely proportional to the time \(t\) taken for effusion. For \(He\), \(r_{He}=\frac{n}{t_{He}}\) (where \(n\) is the number of moles, same for both gases). For the unknown gas, \(r_{unk}=\frac{n}{t_{unk}}\).

Step2: Apply Graham's law of effusion

Graham's law states that \(\frac{r_{He}}{r_{unk}}=\sqrt{\frac{M_{unk}}{M_{He}}}\). Substituting \(r_{He}=\frac{n}{5.06}\) and \(r_{unk}=\frac{n}{17.5}\), we get \(\frac{\frac{n}{5.06}}{\frac{n}{17.5}}=\sqrt{\frac{M_{unk}}{4.003}}\).
Simplifying the left - hand side: \(\frac{17.5}{5.06}=\sqrt{\frac{M_{unk}}{4.003}}\).

Step3: Solve for \(M_{unk}\)

First, calculate \(\frac{17.5}{5.06}\approx3.4585\). Then, square both sides: \((3.4585)^{2}=\frac{M_{unk}}{4.003}\).
\(M_{unk}=(3.4585)^{2}\times4.003\).
\((3.4585)^{2}=11.961\).
\(M_{unk}=11.961\times4.003\approx47.9\)

Answer:

\(47.9\)