1)
Nồng độ phần trăm
\(C\% = \frac{{{m_{chất{\text{ tan}}}}}}{{{m_{dd}}}}.100\% \)
Nồng độ mol:
\({C_M} = \frac{n}{V}{\text{ (M)}}\)
2)
Ta có:
\({m_{KCl}} = {m_{dd}}.C{\% _{KCl}} = 150.2\% = 3{\text{ gam}}\)
3)
\({m_{dd}} = \frac{{{m_{NaCl}}}}{{C{\% _{NaCl}}}} = \frac{{40}}{{20\% }} = 200{\text{ kg}}\)
\( \to {m_{{H_2}O}} = {m_{dd}} - {m_{NaCl}} = 200 - 40 = 160\;{\text{kg}}\)