Đáp án:
\( C{\% _{N{a_2}C{O_3}}} = 28,6\% \)
\( {C_{M{\text{ N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}}} = 3,7787M\)
Giải thích các bước giải:
\({m_{dd}} = 14,3 + 35,7 = 50{\text{ gam}}\)
\( \to C{\% _{N{a_2}C{O_3}}} = \frac{{14,3}}{{50}} = 28,6\% \)
\({V_{dd}} = 35,7{\text{ ml = 0}}{\text{,0357 lít}}\)
\({n_{N{a_2}C{O_3}}} = \frac{{14,3}}{{23.2 + 60}} = 0,1349{\text{ mol}}\)
\( \to {C_{M{\text{ N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}}} = \frac{{0,1349}}{{0,0357}} = 3,7787M\)