A pre-weighed vessel was filled with oxygen at NTP and weighed. It was then evacuated, filled with ${SO}_{2}$ at the same temperature and pressure and again weighed. The weight of oxygen will be

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$\frac{1}{2}$ that of ${SO}_{2}$

Twice that of ${SO}_{2}$

The same as that of ${SO}_{2}$

$\frac{1}{4}$ that of ${SO}_{2}$

answer is B.

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Detailed Solution

An equal volume of gas contains an equal number of moles at constant temperature and pressure.

So, Moles of $O_{2} =$ Moles of ${SO}_{2}$

$1 mole O_{2} = 32 g$ and $1 mole {SO}_{2} = 64 g$

Hence, weight of $O_{2} = \frac{1}{2}$ that of ${SO}_{2}$ .

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