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A solution of KCl has a density of 1.69 ...

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%`

Text Solution

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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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