Question

1.674×10-4 mol of an unidentified gaseous substance effuses through a tiny hole in 86.6 s. Under identical conditions, 1.715×10-4 mol of argon gas takes 84.5 s to effuse. What is the molar mass of the unidentified substance (in g/mol)?

Answer 1

Answer:

44.06 g/mol

Explanation:

We are given;

• Number of moles of unidentified gas as 1.674×10^-4 mol
• Time of effusion of unidentified gas 86.6 s
• Number of moles of Argon gas as 1.715×10^-4 mol
• Time of effusion of Argon gas is 84.5 s

We are supposed to calculate the molar mass of unidentified gas

Step 1: Calculate the effusion rates of each gas

Effusion rate = Number of moles/time

Effusion rate of unidentified gas (R₁)

 =  1.674×10^-4 mol ÷ 86.6 s

 = 1.933 × 10^-6 mol/s

Effusion rate of Argon gas (R₂)

 = 1.715×10^-4 mol ÷ 84.5 sec

= 2.030 × 10^-6 mol/s

Step 2: Calculate the molar mass of unidentified gas

• Assuming the molar mass of unidentified gas is x;
• We can use the Graham's law of effusion to find x;
• According to Graham's law of diffusion;

$\frac{R_{1}}{R_{2}}}=\frac{\sqrt{MM_{Ar}}}{\sqrt{X}}$

But, Molar mass of Argon is 39.948 g/mol

Therefore;

$\frac{1.933*10^-6mol/s}{2.030*10^-6mol/s}}=\frac{\sqrt{39.948}}{\sqrt{X}}$

$0.9522=\frac{\sqrt{39.948}}{\sqrt{X}}$

Solving for X

x = 44.06 g/mol

Therefore, the molar mass of the identified gas is 44.06 g/mol