1)

a)

Ta có:

\({n_{HCl}} = 5.0,5 = 2,5{\text{ mol}}\)

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

\( \to {m_{dd\;{\text{HCl}}}} = \frac{{91,25}}{{36,5\% }} = 250{\text{ gam}}\)

\( \to {V_{dd\;{\text{HCl}}}} = \frac{{250}}{{1,19}} = 210,084{\text{ ml}}\)

b)

Giả sử có \(m\) gam dung dịch \(HCl\)

\( \to {V_{dd\;{\text{HCl}}}} = \frac{m}{{1,19}}{\text{ ml = }}\frac{m}{{1190}}{\text{ lít}}\)

\({m_{HCl}} = m.37\%  = 0,37m{\text{ gam}}\)

\( \to {n_{HCl}} = \frac{{0,37m}}{{36,5}}\)

\( \to {C_{M{\text{ HCl}}}} = \frac{{\frac{{0,37m}}{{36,5}}}}{{\frac{m}{{1190}}}} = 12,063M\)

c)

Giả sử có \(V\) lít \(H_2SO_4\) 2,5M

\( \to {n_{{H_2}S{O_4}}} = 2,5V{\text{ mol}} \to {{\text{m}}_{{H_2}S{O_4}}} = 2,5V.98 = 245V{\text{ gam}}\)

\({V_{dd}} = V{\text{ }}lít = 1000V{\text{ ml}}\)

\( \to {m_{dd}} = 1000V.1,19 = 1190V{\text{ gam}}\)

\( \to C{\% _{{H_2}S{O_4}}} = \frac{{245V}}{{1190V}} = 20,59\% \)

2)

a)

Gọi khối lượng \(KOH\) cần thêm vào là \(x\) gam

Sau khi thêm vào

\({m_{dd}} = 1200 + x{\text{ gam}}\)

\({m_{KOH}} = x + 1200.12\%  = x + 144{\text{ gam}}\)

\( \to C{\% _{KOH}} = \frac{{x + 144}}{{x + 1200}} = 20\%  \to x = 120{\text{ gam}}\)

Cần thêm vào 120 gam \(KOH\)

b)

2 lít  \(HCl\) 4M chứa

\({n_{HCl}} = 2.4 = 8{\text{ mol}}\)

1 lít \(HCl\) 0,5M chứa

\({n_{HCl}} = 1.0,5 = 0,5{\text{ }}mol\)

Hòa 2 dung dịch với nhau

\({n_{HCl}} = 8 + 0,5 = 8,5{\text{ mol}}\)

\( \to {V_{dd}} = 2 + 1 = 3{\text{ lít}}\)

\( \to {C_{M{\text{ HCl}}}} = \frac{{8,5}}{3} = 2,8333{\text{ M}}\)