The total volume of mixture of 2 g of helium and 7 g of nitrogen under S.T.P. conditions is

22.4 lit

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11.2 lit

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16.8 lit

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5.6 lit

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Solution

The correct option is C
16.8 lit

$\frac{W}{G M W} \times 22.4$

= $(\frac{2}{4} + \frac{7}{28}) \times 22.4 = 16.8 L$

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