7)
\({n_{NaOH}} = 0,25.0,5 = 0,125{\text{ mol}}\)
\( \to {V_{dd{\text{ NaOH 2M}}}} = \frac{{0,125}}{2} = 0,0625{\text{ lit = 62}}{\text{,5 ml}}\)
\( \to {m_{{H_2}O}} = 250 - 62,5 = 187,5{\text{ ml}}\)
8)
Phản ứng xảy ra:
\(NaOH + HCl\xrightarrow{{}}NaCl + {H_2}O\)
Ta có:
\({m_{NaOH}} = 200.20\% = 40{\text{ gam}}\)
\( \to {n_{NaOH}} = {n_{HCl}} = {n_{NaCl}} = \frac{{40}}{{23 + 16 + 1}} = 1{\text{ mol}}\)
BTKL:
\({m_{dd}} = {m_{dd\;NaOH}} + {m_{{\text{dd }}HCl}} = 200 + 100 = 300{\text{ gam}}\)
\({m_{NaCl}} = 1.(23 + 35,5) = 58,5{\text{ gam}}\)
\( \to C{\% _{NaCl}} = \frac{{58,5}}{{300}} = 19,5\% \)
\({m_{HCl}} = 1.(1 + 35,5) = 36,5{\text{ gam}}\)
\( \to C{\% _{HCl}} = \frac{{36,5}}{{100}} = 36,5\% \)
