The total pressure of a mixture of 6.4 grams of oxygen and 5.6 grams of nitrogen present in a 2 litre vessel is 1200 $m m$ . What is the partial pressure ( in $m m$ ) of nitrogen?

1200

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600

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900

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200

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Solution

The correct option is B 600
Given, $6.4 g$ of $O_{2} M O_{2} = \frac{6.4}{32} = 0.2$ $5.6 g$ of $N_{2} M N_{2} = \frac{5.6}{2.8} = 0.2$ $V = 2 l T_{p} = 1200 m m .$
Pressure of $Nitrogen=XN2↓Mole fraction of N2$ $T_{p}$
$P = \frac{0.2}{0.4} \times 1200 = 600$ $m m$ of $N_{2} .$

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The total pressure of a mixture of 6.4 grams of oxygen and 5.6 grams of nitrogen present in a 2 litre vessel is 1200 mm. What is the partial pressure ( in mm ) of nitrogen?

The correct option is B 600Given, 6.4g of O2MO2=6.432=0.2 5.6g of N2MN2=5.62.8=0.2 V=2lTp=1200mm.Pressure of Nitrogen=XN2↓Mole fraction of N2 TpP=0.20.4×1200=600 mm of N2.

The total pressure of a mixture of 6.4 grams of oxygen and 5.6 grams of nitrogen present in a 2 litre vessel is 1200 $m m$ . What is the partial pressure ( in $m m$ ) of nitrogen?

The correct option is B 600
Given, $6.4 g$ of $O_{2} M O_{2} = \frac{6.4}{32} = 0.2$ $5.6 g$ of $N_{2} M N_{2} = \frac{5.6}{2.8} = 0.2$ $V = 2 l T_{p} = 1200 m m .$
Pressure of $Nitrogen=XN2↓Mole fraction of N2$ $T_{p}$
$P = \frac{0.2}{0.4} \times 1200 = 600$ $m m$ of $N_{2} .$