Đáp án + giải thích các bước giải:
$n_{CaCl_2.6H_2O}=\frac{25}{219}=0,114(mol)$
Bảo toàn nguyên tố $Ca:$
$-n_{CaCl_2}=n_{CaCl_2.6H_2O}=0,114(mol)$
Ta có $:d_{H_2O}=1(g/ml) \to m_{H_2O}=300(g)$
$mdd_{CaCl_2}=mct_{CaCl_2.6H_2O}+m_{H_2O}=25+300=325(g)$
$ \to C\%_{CaCl_2}=\frac{0,114.111}{325}.100\%=3,89\%$
Bảo toàn nguyên tố $H:$
$-n_{H_2O}=6.n_{CaCl_2.6H_2O}=6.0,114=0,684(mol)$
$ \to m_{H_2O}=0,684.18=12,312(g)$
Mà $d_{H_2O}=1(g/ml)$
$ \to V_{CaCl_2}= ∑V_{H_2O}=300+12,312=312,312(ml)=0,312312(l)$
$\to CM_{CaCl_2}=\frac{0,114}{0,312312}=0,365(M)$
$-d_{CaCl_2}=\frac{325}{312,312}=1,04(g/ml)$
