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.

\( \to C{\% _{MgS{O_4}}} = \frac{{35}}{{135}} = 25,926\% \)

a.

$n_{HCl}=\frac{3,36}{22,4}=0,15(mol)$ 

$⇒m_{ct..HCl}=0,15.36,5=5,475(g)$

-$m_{dd..HCl}=m_{ct..HCl}+m_{H_2O}$

       $=5,475+100=105,475(g)$

$⇒C$%$HCl=\frac{5,475}{105,475}.100$ %$≈5,19$%

b.

-Bảo toàn nguyên tố $Na$:

$⇒n_{Na_2CO_3}=n_{Na_2CO_3.10H_2O}=\frac{14,3}{286}=0,05(mol)$ 

$⇒m_{Na_2CO_3}=0,05.106=5,3(g)$

-$m_{dd..Na_2CO_3}=m_{Na_2CO_3.10H_2O}+m_{H_2O}$

          $=14,3+100=114,3(g)$

$⇒C$%$Na_2CO_3=\frac{5,3}{114,3}.100$ %$≈4,6$%

c.

-Ở $20^oC$ 

⇒ở 100 g $H_2O$ hòa tan được 35 (g) $MgSO_4$

-$m_{dd..MgSO_4}=m_{MgSO_4}+m_{H_2O}$

          $=35+100=135(g)$

$⇒C$%$MgSO_4=\frac{35}{135}.100$ %$≈25,9$%