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The addition of 3 gms of a substance to 100 gms C Cl_(4) (M.w. = 154 gm/mole) raises the boiling point of C Cl_(4) by 0.60^(@)C. If K_(b) of C Cl_(4) is 5.03 K Kg "mole"^(-1) then find out the relative lowering of vapour pressure. |
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Answer» `0.0181` Mol. Wt. of sub `= M_(2)=?` SOLVENT `= C Cl_(4)=100 gms = W_(1)` Molal of solvent `C Cl_(4)=153" gm mole"^(-1)=M_(1)` Increase of B.P. `= Delta T_(b)=0.6^(@)C` `K_(b)` of solvent `C Cl_(4)=5.03" K kg mole"^(-1)` `M_(2)=(K_(b)xx W_(2)xx 100)/(Delta T_(b)xx W_(1))` `therefore M_(2) = (5.03xx3xx1000)/(0.6xx100)` `= 251.5"gm mol"^(-1)` Relative depression of vapour pressure `= (p^(@)-p)/(p^(@))=X_(2)` `= (n_(2))/(n_(1)+n_(2))` Where `n_(2)=` mole of solvent `= (W_(2))/(M_(2))=(3)/(251.5)` = 0.0119 mole substance `n_(1)=` mole of solvent `= (W_(1))/(M_(1))=(100)/(154)` = 0.6494 mole solvent `C Cl_(4)` TOTAL mole `= 0.6494+0.0119=0.6613` `X_(2)=(0.0119)/(0.6613)` `=0.01799 ~~0.0181` |
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