$0.2 g$ of gas $X$ occupies a volume of $440 m l$ and $0.1 g$ of carbon dioxide occupies $320 m l$ at the same temperature and pressure. The gas $X$ could be:

O_{2}

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N O

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{S O}_{2}

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C_{4} H_{10}

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Solution

The correct option is B ${S O}_{2}$
$V \propto n ⟹ \frac{V_{2}}{V_{1}} = \frac{n_{2}}{n_{1}} = c o n s t a n t ⟹ \frac{V_{x}}{V_{{C O}_{2}}} = \frac{n_{x}}{n_{{C O}_{2}}} ⟹ \frac{440 m l}{320 m l} = \frac{\frac{0.2 g}{M_{x}}}{\frac{0.1 g}{44 g {m o l}^{- 1}}} ⟹ M_{x} = 64$
$∴$ The gas will be $S O_{2}$ .

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0.2g of gas X occupies a volume of 440ml and 0.1g of carbon dioxide occupies 320ml at the same temperature and pressure. The gas X could be:

The correct option is B SO2V∝n⟹V2V1=n2n1=constant⟹VxVCO2=nxnCO2⟹440ml320ml=0.2gMx0.1g44gmol−1⟹Mx=64∴ The gas will be SO2.

$0.2 g$ of gas $X$ occupies a volume of $440 m l$ and $0.1 g$ of carbon dioxide occupies $320 m l$ at the same temperature and pressure. The gas $X$ could be: