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How many grams of sucrose (M. wt. = 342) should be dissolved in 100 g water in order to produce a solution with 105.0^(@)C difference between the freezing point and the boiling point ? (K_(f)=1.86^(@)C//m, K_(b)=0.51^(@)C//m) |
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Answer» Solution :`DeltaT_(b)=K_(b)xxm` `"B. PT of solution "(T_(b))=100+DeltaT_(b)=100+K_(b)m` `"F. pt of solution "(T_(b))=0-DeltaT_(f)=0-K_(f)m, T_(b)-T_(f)=(100+K_(b)m)-(-K_(f)m)` `"F. pt. of solution "(T_(f))= 0-DeltaT_(f)=0-K_(f)m,""T_(b)-T_(f)=(100+K_(b)m)-(K_(f)m)` `""105=100+0.51xxm+1.86m.` This gives m = 2.11, i.e., 1000 G of water contain sucrose = 2.11 MOLE `=2.11xx342 g` `therefore"Mass of sucrose to be dissolved in 100 g of water "=(2.1xx342)/(1000)xx100=72g.` |
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