60 cm\(^3\) of SO\(_2\) gas diffused through a porous membrane in \( x \) minutes. Under similar conditions, 360 cm\(^3\) of another gas (molar mass 4 g mol\(^{-1}\)) diffused in \( y \) minutes. The ratio of \( x \) and \( y \) is:
Show Hint
- \( 3:2 \)
- \( 2:3 \)
- \( 1:3 \)
- \( 3:1 \)
The Correct Option is B
Solution and Explanation
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}, \]
where:- \( r_1 \) and \( r_2 \) are the rates of diffusion of gases 1 and 2, respectively,
- \( M_1 \) and \( M_2 \) are the molar masses of gases 1 and 2, respectively.
Step 1: Identify the gases and their molar masses
- Gas 1: \( \text{SO}_2 \) (sulfur dioxide), with molar mass \( M_1 = 64 \, \text{g mol}^{-1} \).
- Gas 2: Another gas, with molar mass \( M_2 = 4 \, \text{g mol}^{-1} \).
Step 2: Express the rates of diffusion The rate of diffusion \( r \) is given by the volume of gas diffused per unit time:
\[ r = \frac{V}{t}, \]
where:- \( V \) is the volume of gas diffused,
- \( t \) is the time taken for diffusion.
\[ r_1 = \frac{60 \, \text{cm}^3}{x \, \text{minutes}}. \]
For the other gas:\[ r_2 = \frac{360 \, \text{cm}^3}{y \, \text{minutes}}. \]
Step 3: Apply Graham's Law Using Graham's Law:
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}. \]
Substitute the expressions for \( r_1 \) and \( r_2 \):\[ \frac{\frac{60}{x}}{\frac{360}{y}} = \sqrt{\frac{4}{64}}. \]
Simplify the left-hand side:\[ \frac{60}{x} \cdot \frac{y}{360} = \frac{y}{6x}. \]
Simplify the right-hand side:\[ \sqrt{\frac{4}{64}} = \sqrt{\frac{1}{16}} = \frac{1}{4}. \]
Thus:\[ \frac{y}{6x} = \frac{1}{4}. \]
Step 4: Solve for the ratio \( \frac{x}{y} \) Cross-multiply:
\[ 4y = 6x. \]
Divide both sides by 2:\[ 2y = 3x. \]
Rearrange to find the ratio \( \frac{x}{y} \):\[ \frac{x}{y} = \frac{2}{3}. \]
Final Answer: The ratio of \( x \) to \( y \) is:
\[ \boxed{2:3} \]
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