a)
Ta có:
\({m_{{H_2}S{O_4}}} = 120.15\% = 18{\text{ gam}}\)
\( \to {m_{{H_2}O}} = 120 - 18 = 102{\text{ gam}}\)
Pha 18 gam \(H_2SO_4\) vào 102 gam nước rồi khuấy đều.
b)
Ta có:
\({n_{NaCl}} = \frac{{11,7}}{{23 + 35,5}} = 0,2{\text{ mol}}\)
\( \to {C_{M{\text{ NaCl}}}} = \frac{{{n_{NaCl}}}}{{{V_{dd}}}} = \frac{{0,2}}{{0,4}} = 0,5M\)