Gọi khối lượng KAl(SO4)2 trong 600g dd bão hòa ở 20oC là a (g)
Có: \(S=\dfrac{a}{600-a}.100=6\left(g\right)\)
=> a = \(\dfrac{1800}{53}\) (g)
=> \(m_{H_2O\left(bđ\right)}=600-\dfrac{1800}{53}=\dfrac{30000}{53}\left(g\right)\)
Gọi số mol KAl(SO4)2.12H2O tách ra là x (mol)
=> \(\left\{{}\begin{matrix}n_{KAl\left(SO_4\right)_2\left(tách.ra\right)}=x\left(mol\right)\\n_{H_2O\left(tách.ra\right)}=12x\left(mol\right)\end{matrix}\right.\)
\(m_{KAl\left(SO_4\right)_2\left(còn.lại\right)}=\dfrac{1800}{53}-258x\left(g\right)\)
\(m_{H_2O\left(còn.lại\right)}=\dfrac{30000}{53}-200-18.12x=\dfrac{19400}{53}-216x\left(g\right)\)
=> \(S=\dfrac{\dfrac{1800}{53}-258x}{\dfrac{19400}{53}-216x}.100=6\left(g\right)\)
=> x = \(\dfrac{50}{1021}\) (mol)
=> \(m_{KAl\left(SO_4\right)_2.12H_2O}=\dfrac{50}{1021}.474=23,2125\left(g\right)\)
