Q1 / 180 · COMEDK 2023 (Evening Shift) · Chemistry
$5.8 \mathrm{~g}$ of a gas maintained at $95^{\circ} \mathrm{C}$ occupies the same volume as $0.368 \mathrm{~g}$ of hydrogen gas maintained at a temperature of $17^{\circ} \mathrm{C}$ and pressure being the same atmospheric pressure for both the gases. What is the molecular mass of the unknown gas?
- Correct answerA40 g/mol
- B71 g/mol
- C44 g/mol
- D32 g/mol
Solution
Using the ideal gas equation $PV = nRT$, where $n = \dfrac{w}{M}$, we have $V = \dfrac{wRT}{PM}$.
Since the volume $V$ and pressure $P$ are the same for both gases, we can write the equality:
$\dfrac{w_1 R T_1}{M_1 P} = \dfrac{w_2 R T_2}{M_2 P}$
Canceling $R$ and $P$ from both sides, we get:
$\dfrac{w_1 T_1}{M_1} = \dfrac{w_2 T_2}{M_2}$
Given values for the unknown gas: $w_1 = 5.8 \text{ g}$, $T_1 = 95 + 273 = 368 \text{ K}$.
Given values for hydrogen gas ($H_2$): $w_2 = 0.368 \text{ g}$, $T_2 = 17 + 273 = 290 \text{ K}$, $M_2 = 2 \text{ g/mol}$.
Substituting these values into the equation:
$\dfrac{5.8 \times 368}{M_1} = \dfrac{0.368 \times 290}{2}$
Solving for $M_1$:
$M_1 = \dfrac{5.8 \times 368 \times 2}{0.368 \times 290}$
$M_1 = \dfrac{5.8 \times 736}{106.72}$
$M_1 = \dfrac{4268.8}{106.72} = 40 \text{ g/mol}$
Answer: 40 g/mol
Since the volume $V$ and pressure $P$ are the same for both gases, we can write the equality:
$\dfrac{w_1 R T_1}{M_1 P} = \dfrac{w_2 R T_2}{M_2 P}$
Canceling $R$ and $P$ from both sides, we get:
$\dfrac{w_1 T_1}{M_1} = \dfrac{w_2 T_2}{M_2}$
Given values for the unknown gas: $w_1 = 5.8 \text{ g}$, $T_1 = 95 + 273 = 368 \text{ K}$.
Given values for hydrogen gas ($H_2$): $w_2 = 0.368 \text{ g}$, $T_2 = 17 + 273 = 290 \text{ K}$, $M_2 = 2 \text{ g/mol}$.
Substituting these values into the equation:
$\dfrac{5.8 \times 368}{M_1} = \dfrac{0.368 \times 290}{2}$
Solving for $M_1$:
$M_1 = \dfrac{5.8 \times 368 \times 2}{0.368 \times 290}$
$M_1 = \dfrac{5.8 \times 736}{106.72}$
$M_1 = \dfrac{4268.8}{106.72} = 40 \text{ g/mol}$
Answer: 40 g/mol
Hence, the correct answer is option: 'A'.