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A solution of `KCl` has a density of `1.69 g mL^(-1)` and is 67% by weight. Find the denisty of the solution if it is diluted so that the percentage by weight of `KCl` in the diluted solution is 30%` |
Answer» Let the volume of the `KCl` solution be `100 mL`. Weight of `KCl` solution `= 100 xx 1.69 = 169 g` `100 g` of solution contains `= 67 g` of `KCl` `169 g` of solution `= (67)/(100) xx 169 = 113.23 g`. Let `x mL` of `H_(2) O` be added. New volume fo soltuion `= (100 + x) mL` New weight of solution `= (169 + x) g` (since `x mL of H_(2)O = x of H_(2) O, d_(H_(2)O) = 1)` New percentage of the solution = 30% % be weight ` = ("Weight of solute" xx 100)/("Weight of solution")` `30 = (113.23)/((169 + x)) xx 100` `X = 208.43 mL = 208.43 g` New density `= ("New weight fo solution")/("New volume of Solution")` `= ((169 + x))/((100+ x))` `((169 + 208.43))/((100 + 208.43)) = (377.43)/(308.43)` `:. d = 1.224` |
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