\(n_{MgSO_4\left(khan\right)}=\dfrac{1,58}{120}=\dfrac{79}{6000}\left(mol\right)\)
=> \(n_{MgSO_4.nH_2O\left(kết.tinh\right)}=\dfrac{79}{6000}\left(mol\right)\)
=> \(\left\{{}\begin{matrix}n_{MgSO_4\left(kết.tinh\right)}=\dfrac{79}{6000}\left(mol\right)\\n_{H_2O\left(kết.tinh\right)}=\dfrac{79n}{6000}\left(mol\right)\end{matrix}\right.\)
Gọi khối lượng MgSO4 trong dd ban đầu là a (g)
\(S=\dfrac{a}{100-a}.100=35,1\left(g\right)\)
=> a = 25,981 (g)
=> mH2O(dd bđ) = 100 - 25,981 = 74,019 (g)
\(m_{MgSO_4\left(dd.sau.khi.thêm\right)}=25,981+1-1,58=25,401\left(g\right)\)
\(m_{H_2O\left(dd.sau.khi.thêm\right)}=74,019-18.\dfrac{79n}{6000}=74,019-0,237n\left(g\right)\)
\(S=\dfrac{25,401}{74,019-0,237n}.100=35,1\left(g\right)\)
=> n = 7
=> CTHH: MgSO4.7H2O
