Đáp án:

 a) Đổi: `500ml=0,5l`

`n_(H_2SO4)=C_(M).V_(dd)=1,25.0,5=0,625 (mol)`

`→m_(H_2SO4)=0,625.98=61,25 (g)`

`m_(dd)=V_(dd).D=500.1,25=625 (g)`

`→C%_(H_2SO_4)=(61,25)/(625).100%=9,8%`

b) `m_(H_2SO4)=400.(2,5%)/(100%)=10 (g)`

`→n_(H_2SO4)=10/98=5/49 (mol)`

`V_(dd)=400/(1,5)=800/3 (ml)=4/15` (lít)

`→C_(M_(H_2SO_4))=5/49 : 4/15≈0,38 (M)`

Em tham khảo nha:

\(\begin{array}{l}
a)\\
{n_{{H_2}S{O_4}}} = 0,5 \times 1,25 = 0,625\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 500 \times 1,25 = 625g\\
{C_\% }{H_2}S{O_4} = \dfrac{{0,625 \times 98}}{{625}} \times 100\%  = 9,8\% \\
b)\\
{n_{{H_2}S{O_4}}} = \dfrac{{400 \times 2,5\% }}{{98}} = \dfrac{5}{{49}}\,mol\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{400}}{{1,5}} = \dfrac{{800}}{3}\,ml = \dfrac{4}{{15}}l\\
{C_M}{H_2}S{O_4} = \dfrac{{\dfrac{5}{{49}}\,}}{{\dfrac{4}{{15}}}} \approx 0,38M
\end{array}\)