InterviewSolution
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InterviewSolution
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`100 mL` of hydrogen was confined in a diffusion tube and exposed to air, and at equilibrium, a volume of `26.1 mL` of air was measured in the tube. Again, when `100 mL` of `CO_(2)` was placed in the same tube and exposed to air, `123 mL` of air was measured in the tube at the equilibrium. Find the molecular weight of `CO_(2)`. |
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Answer» In the first case, when `100 mL` of `H_(2)` is diffused, the volume of air diffused was `26.1 mL`. Thus `(Rate of diffusion of H_(2))/(Rate of diffusion of air)=(100)/(26.1)` …(`i`) `(Rate of diffusion of CO_(2))/(Rate of diffusion of air)=(100)/(123)` ...(`ii`) From equation (`i`) and (`ii`), we get `(r_(H_(2)))/(r_(CO_(2)))=(123)/(261)=sqrt((M_(CO_(2)))/(M_(H_(2))))` `Molecular of weight of CO_(2)=(123xx123)/(26.1xx26.1)=2=44.42` |
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