1)

\({n_{NaOH}} = 0,05.1 = 0,05{\text{ mol;}}{{\text{n}}_{KOH}} = 0,15.0,5 = 0,075{\text{ mol;}}{{\text{V}}_{NaOH}} = 0,05 + 0,15 = 0,2{\text{ lít}}\)

\( \to {C_{M{\text{ NaOH}}}} = \frac{{0,05}}{{0,2}} = 0,25M;{C_{M{\text{ KOH}}}} = \frac{{0,075}}{{0,2}} = 0,375M\)

2)

\({m_{KOH}} = 50.10\%  = 5{\text{ gam;}}{{\text{m}}_{KCl}} = 150.15\%  = 22,5{\text{ gam;}}{{\text{m}}_{dd{\text{ mới}}}} = 50 + 150 = 200{\text{ gam}}\)

\( \to C{\% _{NaOH}} = \frac{5}{{200}}.100\%  = 2,5\% ;C{\% _{KCl}} = \frac{{22,5}}{{200}}.100\%  = 11,25\% \)

Đáp án:

 dưới nha bn

Giải thích các bước giải:

 C1:

nNaOH=0,05 mol

n KOH=0,075 mol

CM NaOH=$\frac{0,05}{0,05+0,15}$ =0,25 M

CMKOH=$\frac{0,075}{0,05+0,15}$ =0,375 M

C2:

m ct KOH= 5 g

m ct KCl=22,5 g

mdd=200g

C%KCl=$\frac{22,5 .100}{200}$ =11,25%

C% NaOH=$\frac{5 .100}{200}$ =2,5%

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