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The vapour pressure of pure liquid solvent A is `0.80`atm.When a non volatile substances B is added to the solvent, its vapour pressure drops to `0.60`atm. Mole fraction of the components B in the solution is:A. `0.50`B. `0.25`C. `0.75`D. `0.40` |
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Answer» Correct Answer - B Acc. To R.L.V.P. `(cancelOP)/(P_(0))=X_(cancelB)` `X_(B)(0.80-0.60)/(0.80)=(0.20)/(0.80)=0.25.` |
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