One mole of a light gas (Mol. mass $= 32$ ) occupies a volume of $20$ lit under certain conditions. Under the same conditions, two moles of a heavy gas (Mol. mass $= 64)$ occupies a volume of :

80 l i t

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40 l i t

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10 l i t

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20 l i t

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Solution

The correct option is B $40 l i t$
Moles of gas $\times$ volume of gas
$\begin{matrix} \frac{m_{1}}{v_{1}} = \frac{m_{2}}{v_{2}} \Rightarrow v_{2} = \frac{n_{2}}{n_{1}} \times v_{1} \Rightarrow \frac{2}{1} \times 20 (L) \Rightarrow 40 L v_{2} = 40 L \end{matrix}$
Option B is correct.

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One mole of a light gas (Mol. mass =32) occupies a volume of 20 lit under certain conditions. Under the same conditions, two moles of a heavy gas (Mol. mass =64) occupies a volume of :

The correct option is B 40 litMoles of gas × volume of gasm1v1=m2v2⇒v2=n2n1×v1⇒21×20(L)⇒40Lv2=40LOption B is correct.

One mole of a light gas (Mol. mass $= 32$ ) occupies a volume of $20$ lit under certain conditions. Under the same conditions, two moles of a heavy gas (Mol. mass $= 64)$ occupies a volume of :

The correct option is B $40 l i t$
Moles of gas $\times$ volume of gas
$\begin{matrix} \frac{m_{1}}{v_{1}} = \frac{m_{2}}{v_{2}} \Rightarrow v_{2} = \frac{n_{2}}{n_{1}} \times v_{1} \Rightarrow \frac{2}{1} \times 20 (L) \Rightarrow 40 L v_{2} = 40 L \end{matrix}$
Option B is correct.