a)
Cho 100g\ dd\ KBr
->m_{KBr}=100.37,5\%=37,5(g)
->m_{H_2O}=100-37,5=62,5(g)
->S_{KBr}^{10^oC}={37,5}/{62,5}.100=60(g)
b)
Ở 10^oC:
S_{KBr}={m_{KBr}}/{m_{H_2O}}.100=60
->m_{KBr}=0,6m_{H_2O}
Mà m_{KBr}+m_{H_2O}=120
->m_{KBr}=45(g);m_{H_2O}=75(g)
Ở 80^oC:
S_{KBr}={m_{KBr}}/{m_{H_2O}}.100=95
->m_{KBr}={95}/{100}.m_{H_2O}=0,95.75=71,25(g)
->m_{KBr(them)}=71,25-45=26,25(g)
