3-7 g of a gas at 25- occupied the same volume as 0-184 g of hydrogen at 17-C and at the same pressure- What is the molecular mass of the gas-

For hydrogen, w = 0.184 g T = 17 + 273 = 290 K M = 2 We know that, PV = wMRT = 0.1842× R × 290 ... (i) For unknown gas, w = 3.7 g ...

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Byju's Answer

Standard XII

Chemistry

Avogadro's Law & Gay Lussac's Law

3.7 g of a ga...

Question

$3.7 g$ of a gas at $25^{\circ}$ occupied the same volume as $0.184 g$ of hydrogen at $17^{\circ} C$ and at the same pressure. What is the molecular mass of the gas?

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Solution

For hydrogen,
$w = 0.184 g$
$T = 17 + 273 = 290 K$
$M = 2$
We know that, $P V = \frac{w}{M} R T$
$= \frac{0.184}{2} \times R \times 290 . . . (i)$
For unknown gas,
$w = 3.7 g$
$T = 25 + 273 = 298 K$
$M = ?$ $P V = \frac{3.7}{M} \times R \times 298 . . . (i i)$
Equating both the equations, we get
$\frac{3.7}{M} \times R \times 298 = \frac{0.184}{2} \times R \times 290$
or $M = \frac{3.7 \times 298 \times 2}{0.184 \times 290} = 41.33$

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3.7 g of a gas at 25∘ occupied the same volume as 0.184 g of hydrogen at 17∘C and at the same pressure. What is the molecular mass of the gas?

For hydrogen,w=0.184 gT=17+273=290 KM=2We know that, PV=wMRT=0.1842×R×290...(i)For unknown gas,w=3.7 gT=25+273=298 KM=?PV=3.7M×R×298...(ii)Equating both the equations, we get3.7M×R×298=0.1842×R×290or M=3.7×298×20.184×290=41.33

$3.7 g$ of a gas at $25^{\circ}$ occupied the same volume as $0.184 g$ of hydrogen at $17^{\circ} C$ and at the same pressure. What is the molecular mass of the gas?