\(m_{CuSO_4(bđ)}=50.5\%=2,5(g)\)

\(m_{H_2O(bđ)}=50-2,5=47,5(g)\)

HS1:

\(m_{H_2O(sau)}=\frac{47,5}{2}=23,75(g)\)

\(C\%_{CuSO_4(sau)}=\frac{2,5}{2,5+23,75}.100\%≈9,52\%\)

\(→\) HS1 sai.

HS2:

\(C\%_{CuSO_4(sau)}=\frac{2,5+2,78}{2,78+50}.100\%≈10,004\%\)

\(→\) HS2 sai.

HS3:

\(n_{CuSO_4(t.the)}=n_{CuSO_4.5H_2O}=\frac{4,63}{250}=0,01852(mol)\)

\(m_{CuSO_4(t.the)}=0,01852.160=2,9632(g)\)

\(C\%_{CuSO_4(sau)}=\frac{2,5+2,9632}{2,9632+50}.100\%≈10,315\%\)

\(→\) HS3 sai.

HS4:

\(m_{CuSO_4\,15\%}=50.15\%=7,5(g)\)

\(C\%_{CuSO_4(sau)}=\frac{7,5+2,5}{50+50}.100\%=10\%\)

Vậy HS4 đúng.

Đáp án:

`HS4` đã làm đúng cách để gia tăng nồng độ lên gấp 2 lần

Giải thích:

`m_{CuSO_{4}\text{5%}}=(m_{dd} × \text{C%})/(100\text{%})=(50 × 5)/(100)= 2,5 (g)`

`m_{CuSO_{4}\text{15%})=(m_{dd} × \text{C%})/(100\text{%})=(50 × 15)/(100)= 7,5 (g)`

`-> ∑m_{CuSO_{4}}=2,5 + 7,5= 10 (g)`

`m_{\text{dd thu được}}=50 + 50= 100 (g)`

`->` $\text{C%}_{CuSO_{4}}=\text{$\dfrac{m_{ct}}{m_{dd}}$ × 100%=$\dfrac{10}{100}$ × 100%= 10%}$