Question:
3 g of $\text{O}_2$ diffuses from a container in 15 min. The mass (in g) of $\text{SO}_2$ diffusing from the same container under the same conditions is (At. wt: $\text{O} = 16$ u, $\text{S} = 32$ u)
3 g of $\text{O}_2$ diffuses from a container in 15 min. The mass (in g) of $\text{SO}_2$ diffusing from the same container under the same conditions is (At. wt: $\text{O} = 16$ u, $\text{S} = 32$ u)
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When using Graham's Law for mass instead of moles, remember the simple rule:
$\frac{w_1}{w_2} = \sqrt{\frac{M_1}{M_2}}$.
This direct relation saves time and avoids conversions into moles first.
$\frac{w_1}{w_2} = \sqrt{\frac{M_1}{M_2}}$.
This direct relation saves time and avoids conversions into moles first.
Updated On: Jul 22, 2026
- $\sqrt{3}$
- $\sqrt{3} \times 2$
- $\sqrt{2} \times 3$
- $\sqrt{2}$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
This question is based on Graham's law of diffusion.
We need to find the mass of $\text{SO}_2$ gas that diffuses under the same conditions of temperature, pressure, and time as $3\text{ g}$ of $\text{O}_2$.
Step 2: Key Formula or Approach:
According to Graham's law of diffusion, the rate of diffusion ($r$) of a gas is inversely proportional to the square root of its molar mass ($M$):
\[ r \propto \frac{1}{\sqrt{M}} \] The rate of diffusion can be written as the number of moles ($n$) diffusing per unit time ($t$):
\[ r = \frac{n}{t} = \frac{w}{M \cdot t} \] where $w$ is the mass of the gas diffused.
Under identical conditions of temperature, pressure, and time ($t_1 = t_2$):
\[ \frac{w_1 / M_1}{w_2 / M_2} = \sqrt{\frac{M_2}{M_1}} \] Simplifying this relation gives:
\[ \frac{w_1}{w_2} = \sqrt{\frac{M_1}{M_2}} \]
Step 3: Detailed Explanation:
• Let Gas 1 be $\text{O}_2$:
Molar mass of $\text{O}_2$ ($M_1$) = $2 \times 16 = 32\text{ g/mol}$.
Given mass ($w_1$) = $3\text{ g}$.
• Let Gas 2 be $\text{SO}_2$:
Molar mass of $\text{SO}_2$ ($M_2$) = $32 + (2 \times 16) = 64\text{ g/mol}$.
Let its mass be $w_2$.
• Applying the simplified equation:
\[ \frac{w_1}{w_2} = \sqrt{\frac{M_1}{M_2}} \] \[ \frac{3}{w_2} = \sqrt{\frac{32}{64}} \] \[ \frac{3}{w_2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]
• Solving for $w_2$:
\[ w_2 = 3\sqrt{2}\text{ g} = \sqrt{2} \times 3\text{ g} \]
Step 4: Final Answer:
The mass of $\text{SO}_2$ diffusing under the same conditions is $\sqrt{2} \times 3\text{ g}$.
This question is based on Graham's law of diffusion.
We need to find the mass of $\text{SO}_2$ gas that diffuses under the same conditions of temperature, pressure, and time as $3\text{ g}$ of $\text{O}_2$.
Step 2: Key Formula or Approach:
According to Graham's law of diffusion, the rate of diffusion ($r$) of a gas is inversely proportional to the square root of its molar mass ($M$):
\[ r \propto \frac{1}{\sqrt{M}} \] The rate of diffusion can be written as the number of moles ($n$) diffusing per unit time ($t$):
\[ r = \frac{n}{t} = \frac{w}{M \cdot t} \] where $w$ is the mass of the gas diffused.
Under identical conditions of temperature, pressure, and time ($t_1 = t_2$):
\[ \frac{w_1 / M_1}{w_2 / M_2} = \sqrt{\frac{M_2}{M_1}} \] Simplifying this relation gives:
\[ \frac{w_1}{w_2} = \sqrt{\frac{M_1}{M_2}} \]
Step 3: Detailed Explanation:
• Let Gas 1 be $\text{O}_2$:
Molar mass of $\text{O}_2$ ($M_1$) = $2 \times 16 = 32\text{ g/mol}$.
Given mass ($w_1$) = $3\text{ g}$.
• Let Gas 2 be $\text{SO}_2$:
Molar mass of $\text{SO}_2$ ($M_2$) = $32 + (2 \times 16) = 64\text{ g/mol}$.
Let its mass be $w_2$.
• Applying the simplified equation:
\[ \frac{w_1}{w_2} = \sqrt{\frac{M_1}{M_2}} \] \[ \frac{3}{w_2} = \sqrt{\frac{32}{64}} \] \[ \frac{3}{w_2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]
• Solving for $w_2$:
\[ w_2 = 3\sqrt{2}\text{ g} = \sqrt{2} \times 3\text{ g} \]
Step 4: Final Answer:
The mass of $\text{SO}_2$ diffusing under the same conditions is $\sqrt{2} \times 3\text{ g}$.
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