1)

Ta có:

\({m_{HCl}} = 150.2,65\%  = 3,975{\text{ gam}}\)

\( \to {n_{HCl}} = \frac{{3,975}}{{36,5}} = 0,1089{\text{ mol}}\)

\( \to {C_{M{\text{ HCl}}}} = \frac{{{n_{HCl}}}}{{{V_{dd}}}} = \frac{{0,1089}}{2} = 0,05445{\text{M}}\)

2)

Ta có:

\({m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{{m_{{H_2}S{O_4}}}}}{{C{\% _{{H_2}S{O_4}}}}} = \frac{{2,45}}{{96\% }} = 2,5{\text{ gam}}\)

\( \to {V_{dd}} = \frac{{{m_{dd}}}}{D} = \frac{{2,5}}{{1,84}} = 1,3587{\text{ ml}}\)

3)

Ta có:

\({n_{HCl}} = {V_{dd}}.{C_{M\;{\text{HCl}}}} = 5.0,5 = 2,5{\text{ mol}}\)

\( \to {m_{HCl}} = {n_{HCl}}.{M_{HCl}} = 2,5.36,5 = 91,25{\text{ gam}}\)

\( \to {m_{dd\;{\text{HCl}}}} = \frac{{{m_{HCl}}}}{{C{\% _{HCl}}}} = \frac{{91,25}}{{36\% }} = 253,472{\text{ gam}}\)

\( \to {V_{dd}} = \frac{{{m_{dd}}}}{D} = \frac{{253,472}}{{1,19}} = 213{\text{ ml}}\)