Question

360 cm$^3$ of a hydrocarbon diffuses in 30 minutes, while under the same conditions 360 cm$^3$ of SO$_2$ gas diffuses in one hour. The molecular formula of the hydrocarbon is

• A. CH$_4$
• B. C$_2$H$_6$
• C. C$_2$H$_4$
• D. C$_2$H$_2$

Hint:

Using a simple frame or just bolding for the box Key Points: Graham's Law: $\frac{r_1{r_2 = \sqrt{\frac{M_2{M_1$ (at constant T, P). Rate ($r$) = Volume ($V$) / Time ($t$). Ensure time units are consistent (e.g., both in minutes). Calculate the unknown molar mass and match it with the options. Molar Mass: SO$_2$ = 64 g/mol, CH$_4$ = 16 g/mol, C$_2$H$_6$ = 30 g/mol, C$_2$H$_4$ = 28 g/mol, C$_2$H$_2$ = 26 g/mol.

Answer 1

The Correct Option is A

This problem utilizes Graham's Law of Diffusion, stating that diffusion (or effusion) rate is inversely proportional to the square root of molar mass (M) at constant temperature and pressure. \[ r \propto \frac{1}{\sqrt{M}} \] Comparing two gases (1 = Hydrocarbon (HC), 2 = SO₂): \[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \quad \Rightarrow \quad \frac{r_{HC}}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{HC}}} \] Diffusion rate (\(r\)) is volume diffused (\(V\)) per unit time (\(t\)).

• Hydrocarbon (HC): \(V_{HC} = 360\) cm³, \(t_{HC} = 30\) minutes.
  \(r_{HC} = \frac{360 \text{ cm}^3}{30 \text{ min}} = 12\) cm³/min.
• Sulfur Dioxide (SO₂): \(V_{SO_2} = 360\) cm³, \(t_{SO_2} = 1\) hour = 60 minutes.
  \(r_{SO_2} = \frac{360 \text{ cm}^3}{60 \text{ min}} = 6\) cm³/min.

Substitute rates into Graham's Law: \[ \frac{r_{HC}}{r_{SO_2}} = \frac{12}{6} = 2 \] Therefore, \[ 2 = \sqrt{\frac{M_{SO_2}}{M_{HC}}} \] Determine the molar mass of SO₂: M(S) \(\approx\) 32 g/mol, M(O) \(\approx\) 16 g/mol. \[ M_{SO_2} = 32 + 2(16) = 32 + 32 = 64 \text{ g/mol} \] Substitute \(M_{SO_2}\) into the equation: \[ 2 = \sqrt{\frac{64}{M_{HC}}} \] Square both sides: \[ 4 = \frac{64}{M_{HC}} \] Solve for the hydrocarbon's molar mass (\(M_{HC}\)): \[ M_{HC} = \frac{64}{4} = 16 \text{ g/mol} \] Identify the hydrocarbon with a 16 g/mol molar mass from the options:

• (A) CH₄: M = 12 + 4(1) = 16 g/mol
• (B) C₂H₆: M = 2(12) + 6(1) = 24 + 6 = 30 g/mol
• (C) C₂H₄: M = 2(12) + 4(1) = 24 + 4 = 28 g/mol
• (D) C₂H₂: M = 2(12) + 2(1) = 24 + 2 = 26 g/mol

The calculated molar mass (16 g/mol) matches Methane (CH₄), which is option (A).