Em tham khảo nha:

\(\begin{array}{l}
a)\\
{M_{CuS{O_4}.10{H_2}O}} = 64 + 32 + 16 \times 4 + 10 \times (2 \times 1 + 16) = 340g/mol\\
\% {m_{{H_2}O}} = \dfrac{{10 \times 18}}{{340}} \times 100\%  = 52,94\% \\
b)\\
{M_{CaS{O_4}.2{H_2}O}} = 40 + 32 + 16 \times 4 + 2 \times (2 \times 1 + 16) = 172g/mol\\
\% {m_{{H_2}O}} = \dfrac{{2 \times 18}}{{172}} \times 100\%  = 20,93\% \\
c)\\
{M_{{K_2}S{O_4}.A{l_2}{{(S{O_4})}_3}.24{H_2}O}} = 39 \times 2 + 32 + 16 \times 4 + 27 \times 2 + 96 \times 3 + 24 \times 18 = 948g/mol\\
\% {m_{{H_2}O}} = \dfrac{{24 \times 18}}{{948}} \times 100\%  = 45,57\% 
\end{array}\)

Đáp án:

$\downarrow$

Giải thích các bước giải:

 a, $\text{$M_{CuSO_4.10H_2O}$ = 64 + 32 + 16.4 + 10. (2 + 16) = 340 g/mol}$

$\rightarrow$ $\text{%$_{H_2O}$ = $\dfrac{10\times18}{340}$. 100% = 52,94%}$

b, $\text{$M_{CaSO_4.2H_2O}$ = 40 + 32 + 16.4 + 2. (2 + 16) = 172 gam}$

$\rightarrow$ $\text{%$_{H_2O}$ = $\dfrac{2\times18}{172}$. 100% = 20,93%}$

c, $\text{$M_{K_2SO_4.Al_2(SO_4)_3.24H_2O}$ = 39.2 + 32 + 16.4 + 27.2 + 3.(32 + 16.4)}$

$\text{ + 24. (2 + 16) = 948 g/mol}$

$\rightarrow$ $\text{%$_{H_2O}$ = $\dfrac{24\times18}{948}$. 100% = 45,57%}$

$\text{#Kudo}$