$0.2 g$ of a gas $X$ occupies a volume of 0.44 liter at given pressure and temperature. Under identical conditions of $P$ and $T$ , $0.1 g$ of the $C O_{2}$ gas occupies $0.32 L$ volume. The gas $X$ can be:

O_{2}

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S O_{2}

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C O_{2}

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C_{4} H_{10}

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Solution

The correct option is B $S O_{2}$
$0.1 g$ of $C O_{2} = \frac{0.1}{M}$ moles of $C O_{2}$ occupies $0.32 L$ .

$\frac{0.2}{M}$ moles of $X$ occupies $0.44 L$

$P V = n R T$

$V α n$ ----- [At constant T and P]

$\frac{V_{({C O}_{2})}}{V_{X}} = \frac{n_{{C O}_{2}}}{n_{X}} \Rightarrow \frac{0.32}{0.44} = \frac{0.1 \times M}{44 \times 0.2}$

$\Rightarrow M = 64$

& Molar mass of ${S O}_{2} = 32 + 32 = 64$

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0.2 g of a gas X occupies a volume of 0.44 liter at given pressure and temperature. Under identical conditions of P and T, 0.1 g of the CO2 gas occupies 0.32 L volume. The gas X can be:

The correct option is B SO20.1g of CO2=0.1M moles of CO2 occupies 0.32L.0.2M moles of X occupies 0.44LPV=nRTVαn ----- [At constant T and P] V(CO2)VX=nCO2nX⇒0.320.44=0.1×M44×0.2 ⇒M=64& Molar mass of SO2=32+32=64

$0.2 g$ of a gas $X$ occupies a volume of 0.44 liter at given pressure and temperature. Under identical conditions of $P$ and $T$ , $0.1 g$ of the $C O_{2}$ gas occupies $0.32 L$ volume. The gas $X$ can be: