Đáp án:

 a) 0,6M và 0,5M

b) 7,68%

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
BaC{l_2} + N{a_2}S{O_4} \to BaS{O_4} + 2NaCl\\
{n_{BaS{O_4}}} = \dfrac{{29,125}}{{233}} = 0,125\,mol\\
{n_{N{a_2}S{O_4}}} = {n_{BaS{O_4}}} = 0,125\,mol\\
{m_{NaCl}} = 26,525 - 0,125 \times 142 = 8,775g\\
{n_{NaCl}} = \dfrac{{8,775}}{{58,5}} = 0,15\,mol\\
{C_M}NaCl = \dfrac{{0,15}}{{0,25}} = 0,6M\\
{C_M}N{a_2}S{O_4} = \dfrac{{0,125}}{{0,25}} = 0,5M\\
b)\\
{n_{BaC{l_2}}} = {n_{BaS{O_4}}} = 0,125\,mol\\
{m_{{\rm{dd}}A}} = 250 \times 1,12 = 280g\\
{m_{{\rm{dd}}BaC{l_2}}} = \dfrac{{0,125 \times 208}}{{20\% }} = 130g\\
{m_{{\rm{dd}}spu}} = 280 + 130 - 29,125 = 380,875g\\
{n_{NaCl}} = 0,15 + 0,125 \times 2 = 0,4\,mol\\
{C_\% }NaCl = \dfrac{{0,4 \times 58,5}}{{380,875}} \times 100\%  = 7,68\% 
\end{array}\)