Question:
3.68 g of hydrogen at 17°C occupies the same volume as 58.0 g of another gas X at 95°C. The molar mass of gas X is (g mol\(^{-1}\)):
3.68 g of hydrogen at 17°C occupies the same volume as 58.0 g of another gas X at 95°C. The molar mass of gas X is (g mol\(^{-1}\)):
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For equal volumes at different temperatures, use \(nT = \text{constant}\) when pressure is constant.
Updated On: Jun 19, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Applying ideal gas relation for equal volumes.
Since both gases occupy the same volume at different temperatures and same pressure, we use: \[ \frac{n_1 T_1}{P} = \frac{n_2 T_2}{P} \] So, \[ n_1 T_1 = n_2 T_2 \]
Step 2: Calculating moles of hydrogen gas.
Given mass of hydrogen = 3.68 g and molar mass = 2 g/mol. So, \[ n_1 = \frac{3.68}{2} = 1.84 \text{ mol} \]
Step 3: Converting temperatures to Kelvin.
\[ T_1 = 17 + 273 = 290 K,\quad T_2 = 95 + 273 = 368 K \]
Step 4: Finding moles of gas X.
Using \(n_1 T_1 = n_2 T_2\), \[ n_2 = \frac{n_1 T_1}{T_2} = \frac{1.84 \times 290}{368} \] \[ n_2 \approx 1.45 \text{ mol} \]
Step 5: Calculating molar mass of gas X.
\[ M = \frac{\text{mass}}{\text{moles}} = \frac{58.0}{1.45} \approx 40 \text{ g/mol} \]
Final Answer: \[ \boxed{40} \]
Since both gases occupy the same volume at different temperatures and same pressure, we use: \[ \frac{n_1 T_1}{P} = \frac{n_2 T_2}{P} \] So, \[ n_1 T_1 = n_2 T_2 \]
Step 2: Calculating moles of hydrogen gas.
Given mass of hydrogen = 3.68 g and molar mass = 2 g/mol. So, \[ n_1 = \frac{3.68}{2} = 1.84 \text{ mol} \]
Step 3: Converting temperatures to Kelvin.
\[ T_1 = 17 + 273 = 290 K,\quad T_2 = 95 + 273 = 368 K \]
Step 4: Finding moles of gas X.
Using \(n_1 T_1 = n_2 T_2\), \[ n_2 = \frac{n_1 T_1}{T_2} = \frac{1.84 \times 290}{368} \] \[ n_2 \approx 1.45 \text{ mol} \]
Step 5: Calculating molar mass of gas X.
\[ M = \frac{\text{mass}}{\text{moles}} = \frac{58.0}{1.45} \approx 40 \text{ g/mol} \]
Final Answer: \[ \boxed{40} \]
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