a)
Gọi khối lượng \(NaCl\) là \(x\)
\( \to {m_{dd}} = {m_{NaCl}} + {m_{{H_2}O}} = x + 123{\text{ gam}}\)
\( \to C{\% _{NaCl}} = \frac{{{m_{NaCl}}}}{{{m_{dd}}}} = \frac{x}{{x + 123}} = 18\% \to x = 27{\text{ gam}}\)
b)
Ta có:
\({m_{KCl}} = 250.14,9\% = 37,25{\text{ gam}}\)
\( \to {n_{KCl}} = \frac{{37,25}}{{39 + 35,5}} = 0,5{\text{ mol}}\)
\( \to {C_{M{\text{ KCl}}}} = \frac{{{n_{KCl}}}}{{{V_{dd}}}} = \frac{{0,5}}{4} = 0,125M\)
