A gas mixture contains 4 $g m s$ of hydrogen and 42 $g m s$ of nitrogen. The fraction of the total pressure exerted by nitrogen is:

\frac{1}{2}

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\frac{3}{2}

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\frac{3}{7}

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\frac{4}{7}

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Solution

The correct option is D $\frac{3}{7}$
Moles of hydrogen $H_{2}$ in 4 grams = $\frac{4}{2} = 2 m o l e s$

Moles of NItrogen $N_{2}$ in 42 grams = $\frac{42}{28} = 1.5 m o l e s$
Total number of moles = 2 + 1.5 = 3.5

The fraction of the total pressure exerted by nitrogen = $\frac{moles of nitrogen}{Total number of moles} = \frac{1.5}{3.5} = \frac{3}{7}$

Hence, option $C$ is correct.

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A gas mixture contains 4 gms of hydrogen and 42 gms of nitrogen. The fraction of the total pressure exerted by nitrogen is:

The correct option is D 37Moles of hydrogen H2 in 4 grams =42=2 molesMoles of NItrogen N2 in 42 grams =4228= 1.5 molesTotal number of moles = 2 + 1.5 = 3.5 The fraction of the total pressure exerted by nitrogen = moles of nitrogen Total number of moles =1.53.5 = 37Hence, option C is correct.

A gas mixture contains 4 $g m s$ of hydrogen and 42 $g m s$ of nitrogen. The fraction of the total pressure exerted by nitrogen is: