Ở \(25^oC:S_{KNO_3}=9,25\left(g\right)\)
Hoà tan 9,25 gam KNO3 vào 100 gam nước thì được ddbh
=> Trong 109,25 gam ddbh KNO3 có 9,25 gam KNO3.
=> Trong 500 gam ddbh KNO3 có x gam KNO3.
\(\rightarrow x=\dfrac{9,25.500}{109,25}=\dfrac{18500}{437}\approx42,33\left(g\right)\)
\(\rightarrow m_{H_2O}=500-42,33=457,67\left(g\right)\)
Ở \(60^oC:S_{KNO_3}=32,8\left(g\right)\)
\(\rightarrow\dfrac{m_{KNO_3}}{m_{H_2O}}.100=32,8\left(g\right)\\ \rightarrow m_{KNO_3}=\dfrac{32,8}{100}.457,67=150,11567\left(g\right)\\ \rightarrow m_{KNO_3\left(thêm\right)}=150,11567-42,33=107,78576\left(g\right)\)