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%`

Explanation

Let the volume of the `KCl` solution be `100mL`
Weight of `KCl` solution `= 100 xx 1.69 = 169 g`
100 g of solution contains = 67 g of KCl
`169g` of solution `=(67)/(100) xx 169 = 113.23g`
Lex x mL of `H_(2)O` be added
New volume of solution `= (100 + x) mL`
New weight of solution `= (169 +x)g`
`("Since " x" " mL " of " H_(2)O = x" " g " of " H_(2)O,d_(H_(2_(O))) = 1)`
New percentage of the solution `=30%`
`%` by weight `=("weight of solute" xx 100)/("weight of solution")`
`30 = (113.23)/((169 xx x)) xx 100`
`x = 208.43mL = 208.43g`
New density = `("New weight of solution")/("New volume of solution")`
`=((169 +x))/((100+x))`
`=((169 + 208 .43))/((100 + 208.43))=(377.43)/(308.43)`
`:. d = 1.224` .