Đáp án `+` Giải thích các bước giải:

`a)`

`C%_(CuSO_4) = (30xx100%)/(30+220) = 12%`

`b)`

`m_(NaOH 12%) = (175xx12%)/(100%) = 21` `(g)`

`=>` `m_(text(NaOH sau khi thêm)) = 21 + 25 = 46` `(g)`

`=>` `C%_(text(NaOH sau khi thêm)) = (46xx100%)/(175 + 25) = 23%`

`c)`

`m_(BaCl_2 10%) = (250xx10%)/(100%) = 25` `(g)`

`m_(BaCl_2 18%) = (150xx18%)/(100%) = 27` `(g)`

`=>` `C%_(text(BaCl2 sau khi trộn)) = (25 + 27)/(250+150)xx100% = 13%`

`d)`

`n_(K_2O) = (4,7)/(94) = 0,05` `(mol)`

`K_2O + H_2O -> 2KOH`

Ta có : `n_(KOH) = 2n_(K_2O) = 0,1` `(mol)`

`=>` `m_(KOH) = 0,1xx56 = 5,6` `(g)`

Mà `m_(dd) = 4,7 + 195,3 = 200` `(g)`

`=>` `C%_(KOH) = (5,6xx100%)/(200%) = 2,8%`

Đáp án:

a) 

`m_{dd}=m_{ct} + m_{dm}=30 + 220= 250 (g)`

`=>` $\text{C%}_{CuSO_{4}}=\text{$\dfrac{m_{ct}}{m_{dd}}$ × 100%=$\dfrac{30}{250}$ × 100%= 12%}$

b)

`m_{dd}=25 + 175= 200 (g)`

`m_{NaOH\text{(12%)}}=(m_{dd} × \text{C%})/(100 \text{%})=(175 × 12 \text{%})/(100 \text{%})= 21 (g)`

`=> ∑m_{NaOH}=25 + 21= 46 (g)`

`=>` $\text{C%}_{NaOH}=\text{$\dfrac{m_{ct}}{m_{dd}}$ × 100%=$\dfrac{46}{200}$ × 100%= 23%}$

c)

`m_{BaCl_{2}\text{(10%)}}=(m_{dd} × \text{C%})/(100 \text{%})=(250 × 10 \text{%})/(100 \text{%})= 25 (g)`

`m_{BaCl_{2}\text{(18%)}}=(m_{dd} × \text{C%})/(100 \text{%})=(150 × 18 \text{%})/(100 \text{%})= 27 (g)`

`=> ∑m_{BaCl_{2}}=25 + 27= 52 (g)`

`m_{dd}=250 + 150= 400 (g)`

`=>` $\text{C%}_{BaCl_{2}}=\text{$\dfrac{m_{ct}}{m_{dd}}$ × 100%=$\dfrac{52}{400}$ × 100%= 13%}$

d)

`K_{2}O + H_{2}O->2KOH`

`n_{K_{2}O}=(m)/(M)=(4,7)/(94)= 0,05 (mol)`

`m_{\text{dd sau pứ}}=4,7 + 195,3= 200 (g)`

Theo PT: `n_{KOH}=2n_{K_{2}O}=2.0,05= 0,1 (mol)`

`=> m_{KOH}=n.M=0,1 × 56= 5,6 (g)`

`=>` $\text{C%}_{KOH}=\text{$\dfrac{m_{ct}}{m_{dd}}$ × 100%=$\dfrac{5,6}{200}$ × 100%= 2,8%}$