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Standard XII

Chemistry

Relating Translational KE and T

Question

$60$ % of $H_{2}$ and $40$ % of $O_{2}$ by volume are present in a mixture at a pressure of $100$ $m m$ . Pressure exerted by $H_{2}$ is:

50

m m

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400

m m

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60

m m

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1000

m m

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Solution

The correct option is C $60$ $m m$
$P_{H_{2}} = P X_{H_{2}}$
Mole fraction of $H_{2} = 0.6$
Thus, partial pressure of $H_{2} = 0.6 \times 100 = 60$ $m m$

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60 % of H2 and 40% of O2 by volume are present in a mixture at a pressure of 100 mm. Pressure exerted by H2 is:

The correct option is C 60 mmPH2=PXH2Mole fraction of H2=0.6Thus, partial pressure of H2=0.6×100=60 mm

$60$ % of $H_{2}$ and $40$ % of $O_{2}$ by volume are present in a mixture at a pressure of $100$ $m m$ . Pressure exerted by $H_{2}$ is:

The correct option is C $60$ $m m$
$P_{H_{2}} = P X_{H_{2}}$
Mole fraction of $H_{2} = 0.6$
Thus, partial pressure of $H_{2} = 0.6 \times 100 = 60$ $m m$