$1$ litre of a mixture of $C O$ and $C O_{2}$ is taken. This mixture is passed through a tube containing red-hot charcoal. The volume now becomes $1.6 l i t e r s$ . The volumes are measured under the same conditions. Find the composition of the mixture by volume.

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Solution

$\begin{matrix} P V = n R T V α \frac{1}{m} H e r e, V = v o l u m e m = m o l e c u l a r w e i g h t ∴ V_{C O} + V_{C O_{2}} = 1.6 L - - - (i) \frac{V_{C O}}{V_{C O_{2}}} = \frac{m_{C O_{2}}}{m_{C O}} = \frac{44}{28} ∴ V_{C O} = \frac{11}{7} V_{C O_{2}} O n p u t t i n g t h e v a l u e o f (1) w e g e t \frac{11}{7} V_{C O_{2}} + V_{C O_{2}} = 1.6 L V_{C O_{2}} = \frac{1.6 \times 7}{18} = \frac{11.2}{18} = 0.62 L a n d V_{C O} = \frac{11}{7} V_{C O_{2}} = 0.98. H e n c e, V_{C O_{2}} = 0.62 L a n d V_{C O} = 0.98 L \end{matrix}$

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1 litre of a mixture of CO and CO2 is taken. This mixture is passed through a tube containing red-hot charcoal. The volume now becomes 1.6 liters. The volumes are measured under the same conditions. Find the composition of the mixture by volume.

PV=nRTVα1mHere,V=volumem=molecularweight∴VCO+VCO2=1.6L−−−(i)VCOVCO2=mCO2mCO=4428∴VCO=117VCO2Onputtingthevalueof(1)weget117VCO2+VCO2=1.6LVCO2=1.6×718=11.218=0.62LandVCO=117VCO2=0.98.Hence,VCO2=0.62LandVCO=0.98L

$1$ litre of a mixture of $C O$ and $C O_{2}$ is taken. This mixture is passed through a tube containing red-hot charcoal. The volume now becomes $1.6 l i t e r s$ . The volumes are measured under the same conditions. Find the composition of the mixture by volume.