Đáp án:
\({{\text{C}}_{M{\text{ CuS}}{{\text{O}}_4}}} = 0,3125M\)
\({\text{\% }}{{\text{m}}_{CuS{O_4}}} = 4,76\% \)
Giải thích các bước giải:
Ta có:
\({n_{CuS{O_4}}} = \frac{{10}}{{160}} = 0,0625{\text{ mol}} \to {{\text{C}}_{M{\text{ CuS}}{{\text{O}}_4}}} = \frac{{0,0625}}{{0,2}} = 0,3125M\)
Ta có:
\({m_{{H_2}O}} = 200.1 = 200{\text{ gam}} \to {{\text{m}}_{dd}} = 200 + 10 = 210{\text{ gam}} \to {\text{\% }}{{\text{m}}_{CuS{O_4}}} = \frac{{10}}{{210}} = 4,76\% \)