Đáp án:
\({m_{dd}} = 200{\text{ gam}}\)
\(C{\% _{NaCl}} = 15\% \)
\({V_{dd}} = 181,81{\text{ ml = 0}}{\text{,18181 lít}}\)
\( {C_{M{\text{ NaCl}}}} = 2,82{\text{ M}}\)
Giải thích các bước giải:
BTKL:
\({m_{dd}} = {m_{NaCl}} + {m_{{H_2}O}} = 30 + 170 = 200{\text{ gam}}\)
\( \to C{\% _{NaCl}} = \frac{{{m_{NaCl}}}}{{{m_{dd}}}} = \frac{{30}}{{200}}.100\% = 15\% \)
\({V_{dd}} = \frac{{{m_{dd}}}}{D} = \frac{{200}}{{1,1}} = 181,81{\text{ ml = 0}}{\text{,18181 lít}}\)
\({n_{NaCl}} = \frac{{30}}{{23 + 35,5}} = 0,513{\text{ mol}}\)
\( \to {C_{M{\text{ NaCl}}}} = \frac{{{n_{NaCl}}}}{{{V_{dd}}}} = \frac{{0,513}}{{0,18181}} = 2,82{\text{ M}}\)