InterviewSolution
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InterviewSolution
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For a dilute solution containing `2.5 g` of a non-volatile non-electrolyte solution in `100g` of water, the elevation in boiling point at `1` atm pressure is `2^(@)C`. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of `Hg)` of the solution is: (take `k_(b) = 0.76 K kg mol^(-1))`A. 724B. 740C. 736D. 718 |
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Answer» Correct Answer - A `M_(2) = (K_(b) xx W_(2))/(W_(1) xx Delta T_(b)) xx 1000 = (1000 xx 0.76 xx 2.5)/(100 xx 2) =9.5` `(p_(1)^(o)-p)/p_(1)^(o) = (W_(2)//M_(2))/(W_(1)//M_(1))` `:. (760-p)/760 = (2.5 // 9.5)/(100//18)=0.047` or 760 - p = `760 xx 0.047` = 35.7 or p=724.3 mm |
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