A sample of water gas contains $42 %$ by volume of carbon monoxide. If the total pressure is $760$ mm of Hg, the partial pressure of carbon monoxide is :

380

mm of Hg

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319.2

mm of Hg

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38

mm of Hg

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360

mm of Hg

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Solution

The correct option is B $319.2$ mm of Hg

$1.$ $42 %$ of CO is released
$100 %$ $-$ 28g
$42 %$ $-$ $x$

$x = \frac{42 \times 28}{100}$
$x = 11.76$

$2.$ for $H_{2} O$
100% $-$ 18g
58% $-$ $x$

$x = \frac{58 \times 16}{100}$

$x = 10.44$
mole of CO $= \frac{11.76}{28} = 0.42$

moles of $H_{2} O$ $= \frac{10.44}{18} = 0.58$

Dalton's law (p) $= P_{T} \times X_{Q}$

$= 760 \times \frac{0.42}{0.42 + 0.58}$

$P = 319.2 m m$

The correct option is B.

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A sample of $O_{2}$ gas is collected over water at $23^{\circ}$ C at a barometric pressure of 751 mm Hg (vapour pressure of water at $23^{\circ}$ C is 21 mm Hg). The partial pressure of $O_{2}$ gas in the sample collected is ......

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A sample of water gas contains 42% by volume of carbon monoxide. If the total pressure is 760 mm of Hg, the partial pressure of carbon monoxide is :

The correct option is B 319.2 mm of Hg1. 42% of CO is released 100% − 28g 42% − x x=42×28100x=11.762. for H2O 100% − 18g 58% − xx=58×16100x=10.44 mole of CO =11.7628=0.42 moles of H2O =10.4418=0.58Dalton's law (p) =PT×XQ =760×0.420.42+0.58 P=319.2mmThe correct option is B.

A sample of water gas contains $42 %$ by volume of carbon monoxide. If the total pressure is $760$ mm of Hg, the partial pressure of carbon monoxide is :