The acid solution has a specific gravity of 1.8, when it contains 62% by weight of the acid. The solution is diluted to such an extant this its specific gravity is lowered to 1.2. what is the % by weight of the acid new solution.

Explanation

Let `100 mL` of the solution is taken
weight of solution `= 100 xx 1.8 = 180 g`
`100 g` fo solution `= 62 g` of acid
`180 g` of solution `= (62)/(100) xx 180 = 111.6 g` of acid
Let `x mL` of `H_(2) O` is added.
`:.` Now volume `= (100 + x) mL`
New weight of solution `= (180 + x) g`
(since `x mL` of `H_(2)O = x g of H_(2)O, d_(H_(2)O) = 1)`
`:.` New density `= (180 + x)/(100 + x)`
`:. 1.2 = (180 + x)/(100 + x)`
Solving for `x`. we get
`:. x = 300 mL = 300 g`
weight of solution `= 180+ 300 = 480 g`
% weight of the acid `= (W_(2))/(W_(sol)) xx 100`
`= (111.6)/(480) xx 100 = 23.25%`