............. .............

9.

`m_{dd\ X}=25+100=125(g)`

`V_{dd\ X}={125}/{1,143}\approx 109,4(ml)`

`C\%_X={25}/{125}.100\%=20\%`

`->` Chọn `D`

10.

Hòa tan tinh thể thu được `A:\ Al_2(SO_4)_3`

`n_{BaSO_4}={0,699}/{233}=0,003(mol)`

`Al_2(SO_4)_3+3BaCl_2->3BaSO_4+2AlCl_3`

Theo PT: `n_{Al_2(SO_4)_3}=1/3n_{BaSO_4}=0,001(mol)`

`->n_{Al_2(SO_4)_3(bd)}=0,001.10=0,01(mol)`

Bảo toàn Al: `n_{Al_2(SO_4)_3.xH_2O}=n_{Al_2(SO_4)_3(bd)}=0,01(mol)`

`->M_{Al_2(SO_4)_3.xH_2O}=342+18x={6,66}/{0,01}=666`

`->x=18`

Vậy CT tinh thể là `Al_2(SO_4)_3.18H_2O`