There are $6.02 \times 10^{22}$ molecules each of $N_{2}, O_{2}$ and $H_{2}$ which are mixed together at $760 m m$ and $273 K$ . The mass of the mixture in grams is:

6.2

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4.12

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3.09

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7

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Solution

The correct option is A $6.2$
$6.02 \times 10^{22}$ molecules of each $N_{2}, O_{2}$ and $H_{2}$

$= \frac{6.02 \times 10^{22}}{6.02 \times 10^{23}}$ moles of each

Weight of mixture = weight of $0.1$ mole $N_{2} +$ weight of $0.1$ mole $H_{2} +$ weight of $0.1$ mole of $O_{2}$

$= (28 \times 0.1) + (2 \times 0.1) + (32 \times 0.1) = 6.2 g m$

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