Ở `50^oC`

`114g ``NaNO_3` + `100g` `H_2O` `= 214g` ddbh

Vậy `m_{NaNO_3,50^oC}` + `m_{H_2O}` = 300g ddbh

`=> m_{NaNO_3,50^oC} = (300 . 114)/(214) = (17100)/(107)g`

`m_{H_2O} = 300 - (17100)/(107) = (15000)/(107)g`

Ở `20^oC:`

`88g` `NaNO_3` hòa tan tối đa `100g` `H_2O`

`m_{NaNO_3,20^oC} `hòa tan tối đa `(15000)/(107)g` `H_2O`

`=> m_{NaNO_3,50^oC} = ((15000)/(107) . 88)/(100) = (13200)/(107)g`

`=> m_{NaNO_3(kt)} = (17100)/(107) - (13200)/(107) ~~ 36,45g`

`50^oC:`

`S_{NaNO_3}={m_{NaNO_3}}/{m_{H_2O}}.100=114(g)`

`->m_{NaNO_3}=1,14m_{H_2O}`

`m_{NaNO_3}+m_{H_2O}=300(g)`

`->m_{NaNO_3}=159,81(g);m_{H_2O}=140,19(g)`

`20^oC:`

`S_{NaNO_3}={m_{NaNO_3}}/{m_{H_2O}}.100=88(g)`

`->m_{NaNO_3}=0,88m_{H_2O}=0,88.140,19=123,3672(g)`

`->m_{NaNO_3\ kt}=159,81-123,3672=36,4428(g)`