ở `20^c` có `35,9g` `NaCl` tan trong 100g nước tạo thành `135,9g` dd
`->` $m_{NaCl20^c}$ `=` $\dfrac{600.35,9}{135,9}$ `=158,5g`
`->` $m_{H_2O}$ `=600 - 158,5 = 441,5g`
Đặt $m_{NaCl}$ cần thêm `=x` mol
`->` $S_{NaCl90^c}$ `=` $\dfrac{158,5+x}{441,5}$ `.100 = 38,5g`
`-> x = 11,5g`

`20^oC:`

`S_{NaCl}={m_{NaCl}}/{m_{H_2O}}.100=35,9(g)`

`->m_{NaCl}=0,359m_{H_2O}(1)`

`m_{NaCl}+m_{H_2O}=600(g)(2)`

`(1)(2)->m_{NaCl}=158,5(g);m_{H_2O}=441,5(g)`

`90^oC:`

`S_{NaCl}={m_{NaCl}}/{m_{H_2O}}.100=38,5(g)`

`->m_{NaCl}=0,385m_{H_2O}=0,385.441,5=169,9775(g)`

`->m_{NaCl\text{ thêm}}=169,9775-158,5=11,4775(g)`