One mole of a light gas (mol. wt. $=$ 32) occupies a volume of $20 L$ under certain conditions. Under the same conditions two moles of a heavy gas (mol. wt. $=$ 64) occupies a volume of:

80 L

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40 L

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10 L

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20 L

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Solution

The correct option is D $40 L$
$V_{1} n_{2} = V_{2} n_{1}$
$20 \times 2 = V_{2} \times 1$
Thus, $V_{2} = 40 L$

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One mole of a light gas (mol. wt. =32) occupies a volume of 20L under certain conditions. Under the same conditions two moles of a heavy gas (mol. wt.= 64) occupies a volume of:

The correct option is D 40LV1n2=V2n120×2=V2×1Thus, V2=40L

One mole of a light gas (mol. wt. $=$ 32) occupies a volume of $20 L$ under certain conditions. Under the same conditions two moles of a heavy gas (mol. wt. $=$ 64) occupies a volume of:

The correct option is D $40 L$
$V_{1} n_{2} = V_{2} n_{1}$
$20 \times 2 = V_{2} \times 1$
Thus, $V_{2} = 40 L$