Đáp án:
\({\text{C}}{{\text{\% }}_{CuS{O_4}}} = 20\% ;{C_{M{\text{ Cu S}}{{\text{O}}_4}}} = 1,67M;{d_{dd}} = 1,333g/ml\)
Giải thích các bước giải:
Ta có:
\({n_{CuS{O_4}.5{H_2}O}} = \frac{{50}}{{160 + 18.5}} = 0,25{\text{ mol = }}{{\text{n}}_{CuS{O_4}}} \to {m_{CuS{O_4}}} = 0,25.160 = 40{\text{ gam}}\)
Ta có:
\({m_{{H_2}O}} = 150.1 = 150{\text{ gam}}\)
BTKL:
\({m_{dd}} = 50 + 150.1 = 200{\text{ gam}} \to {\text{C}}{{\text{\% }}_{CuS{O_4}}} = \frac{{40}}{{200}} = 20\% ;{C_{M{\text{ Cu S}}{{\text{O}}_4}}} = \frac{{0,25}}{{0,15}} = 1,67M;{d_{dd}} = \frac{{200}}{{150}} = 1,333g/ml\)