a)
Ta có: \({n_{FeC{l_3}}} = \frac{{6,5}}{{56 + 35,5.3}} = 0,04{\text{ mol}}\)
BTKL:
\({m_{dd}} = 6,5 + 193,5 = 200{\text{ gam}} \to {{\text{V}}_{dd}} = \frac{{200}}{{1,15}} = 173,913ml = 0,173913{\text{ lít}} \to {{\text{C}}_{M{\text{ FeC}}{{\text{l}}_3}}} = \frac{{0,04}}{{0,173913}} = 0,23M\)
b)
Ta có:
\({n_{NaOH(1)}} = 0,5.1,2 = 0,6{\text{ mol; }}{{\text{n}}_{NaOH(2)}} = 0,3.0,5 = 0,15{\text{ mol}} \to {{\text{n}}_{NaOH}} = 0,6 + 0,15 = 0,75{\text{ mol;}}{{\text{V}}_{dd}} = 500 + 300 = 800{\text{ ml = 0}}{\text{,8 lít}}\)
\( \to {C_{M{\text{ NaOH}}}} = \frac{{0,75}}{{0,8}} = 0,9375M\)
c)
Ta có: \({n_{NaOH}} = 0,125.1 = 0,125{\text{ mol;}}{{\text{V}}_{dd}} = 125 + 75 = 200{\text{ ml = 0}}{\text{,2 lít}} \to {{\text{C}}_{M{\text{ NaOH}}}} = \frac{{0,125}}{{0,2}} = 0,625M\)
d)
Ta có:
\({n_{{H_2}S{O_4}(1)}} = 0,15.2 = 0,3{\text{ mol;}}{{\text{V}}_{{H_2}S{O_4}(2)}} = \frac{{200}}{{1,29}} = 155ml \to {n_{{H_2}S{O_4}(2)}} = 0,155.5 = 0,775{\text{ mol}} \to {{\text{n}}_{{H_2}S{O_4}}} = 0,3 + 0,775 = 1,075{\text{ mol;}}{{\text{V}}_{dd}} = 150 + 155 = 305ml = 0,305 \to {C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{1,075}}{{0,305}} = 3,524M\)
