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Vapour pressure of `C CL_(4)` at `25^@C` is `143` mmHg 0.05g of a non-volatile solute (mol.wt.=`65`)is dissolved in `100ml C CL_(4)`. find the vapour pressure of the solution (density of `C CL_(4)=158g//cm^2`)A. `141.9 mmHg`B. `94.4mmHg`C. `-1.86^(@)C`D. `0.93^(@)C` |
Answer» Correct Answer - A Wt. of `CCl_(4)=(153.8)/(154)approx1` moles of solute `=(0.5)/(65)=0.00769` Now `(P_(0)-P_(S))/(P_(S))=(n)/(N)` or `(143-P_(S))/(P_(S))=(0.00769)/(1)` so `P_(s)=141.9mm Hg`. |
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