Question

1. a sample of oxygen gas was found to effuse at a rate equal to three times that of an unknown gas. the molar mass of the unknown gas is ____ g/mol. (molar mass of o₂ = 32.0 g/mol) a. 288 g/mol b. 96 g/mol c. 55 g/mol d. 4 g/mol e. 10.7 g/mol

Explanation:

Step1: Apply Graham's law of effusion

Graham's law is \( \frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}} \), where \( r \) is the rate of effusion and \( M \) is the molar mass. Let \( r_1 \) be the rate of \( O_2 \), \( M_1 = 32.0\ g/mol \), \( r_2 \) be the rate of the unknown gas, and \( M_2 \) be the molar mass of the unknown gas. Given \( r_1 = 3r_2 \).

Step2: Substitute values into the formula

Substitute \( r_1 = 3r_2 \) into \( \frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}} \). We get \( \frac{3r_2}{r_2}=\sqrt{\frac{M_2}{32.0}} \). Simplify the left - hand side to \( 3=\sqrt{\frac{M_2}{32.0}} \).

Step3: Solve for \( M_2 \)

Square both sides of the equation \( 3=\sqrt{\frac{M_2}{32.0}} \). We have \( 3^2=\frac{M_2}{32.0} \), so \( M_2=9\times32.0 = 288\ g/mol \).

Answer:

A. \( 288\ g/mol \)