Đáp án:

\(C{\% _{CuS{O_4}}} = 8\% \)

\( {C_{M{\text{ CuS}}{{\text{O}}_4}}} = 0,5714{\text{ M}}\)

\({V_{{H_2}O}}  = 72{\text{ ml}}\)

Giải thích các bước giải:

 Ta có:

\({m_{{H_2}O}} = 105.1 = 105{\text{ gam}}\)

\( \to {m_{dd}} = {m_{CuS{O_4}.5{H_2}O}} + {m_{{H_2}O}} = 15 + 105 = 120{\text{ gam}}\)

\( \to {n_{CuS{O_4}.5{H_2}O}} = {n_{CuS{O_4}}} = \frac{{15}}{{160 + 18.5}} = 0,06{\text{ mol}}\)

\( \to {m_{CuS{O_4}}} = 0,06.160 = 9,6{\text{ gam}}\)

\( \to C{\% _{CuS{O_4}}} = \frac{{9,6}}{{120}} = 8\% \)

\({V_{dd}} = {V_{{H_2}O}} = 105{\text{ ml = 0}}{\text{,105 lít}}\)

\( \to {C_{M{\text{ CuS}}{{\text{O}}_4}}} = \frac{{0,06}}{{0,105}} = 0,5714{\text{ M}}\)

Sau khi pha thêm nước dung dịch có nồng độ là 5%.

\( \to {m_{dd}} = \frac{{9,6}}{{5\% }} = 192{\text{ gam}} \to {{\text{m}}_{{H_2}O}} = 192 - 120 = 72{\text{ gam}}\)

\( \to {V_{{H_2}O}} = \frac{{72}}{1} = 72{\text{ ml}}\)