a)
Ta có:
\({n_{HCl}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}}\)
\( \to {m_{HCl}} = 0,15.36,5 = 5,475{\text{ gam}}\)
BTKL:
\({m_{dd}} = {m_{HCl}} + {m_{{H_2}O}} = 5,475 + 100 = 105,475{\text{ gam}}\)
\( \to C{\% _{HCl}} = \frac{{5,475}}{{105,475}} = 5,19\% \)
b)
Ta có:
\({n_{N{a_2}C{O_3}.10{H_2}O}} = \frac{{14,3}}{{23.2 + 60 + 18.10}} = 0,05{\text{ mol = }}{{\text{n}}_{N{a_2}C{O_3}}}\)
\( \to {m_{N{a_2}C{O_3}}} = 0,05.(23.2 + 60) = 5,3{\text{ gam}}\)
\({m_{dd}} = {m_{N{a_2}C{O_3}.10{H_2}O}} + {m_{{H_2}O}} = 100 + 14,3 = 114,3{\text{gam}}\)
\( \to C{\% _{N{a_2}C{O_3}}} = \frac{{5,3}}{{114,3}} = 4,637\% \)
c)
Ở 20 độ C có 35 gam \(MgSO_4\) tan trong 100 gam nước tạo 135 gam dung dịch bão hòa.
\( \to C{\% _{MgS{O_4}}} = \frac{{35}}{{135}} = 25,926\% \)