Đặt `m_{CaCl_2} = x(g)`

`m_{CaCl_2(250g dd CaCl_2 2%)} = 250 . 2% = 5`g

`m_{CaCl_2 sau} = 5 + x(g)`

`m_{dd sau} = 250 + x(g)`

`C%_{dd CaCl_2} = {5 + x}/{250 + x} . 100% = 9%`

`=> x` $\approx$ `19,231`

Vậy cần thêm `19,231g` `CaCl_2` vào `250g` dd `CaCl_2` `2%` để tạo dd `CaCl_2` `9%`

`m_{CaCl_2\ 2\%}={250.2}/{100}=5(g)`

Đặt `m_{CaCl_2}=x(g)`

`->C\%_{CaCl_2\ sau}={x+5}/{x+250}.100\%=9\%`

`->x\approx 19,23(g)`

`->m_{CaCl_2\ them}\approx 19,23(g)`