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Câu 5: `a)` `-` $m_{NaCl}$ `=` $\dfrac{36.100}{100}$ `= 36` `(g)`

`->` `C%_{NaCl} =` `\frac{36 xx 100%}{136} = 26,47%` 

`-` $m_{đường}$ `=` $\dfrac{204.100}{100}$ `= 204` `(g)` 

`->` `C%_{đường} `= `\frac{204 xx 100%}{304}` `= 67,10%` 

`b)` `-` $n_{KOH}$ `=` $\dfrac{5,6}{56}$ `= 0,1` `(mol)` 

`-` $n_{NaCl}$ `=` $\dfrac{5,85}{58,5}$ `= 0,1` `(mol)` 

`-` `150` `(ml)` `= 0,15` `(l)`

`->` `C_{M \ NaCl} =` $C_{M \ {đường}}$ `=` $\dfrac{0,1}{0,15}$ `= 0,67` `(M)` 

Câu `5`:

`a)` `S=\frac{m_{NaCl}}{m_{\text{dm}}}.100`

`=> m_{NaCl}=\frac{36.100}{100}=36g`

Tương tự :`m_{\text{nước đường}}=204g`

`C%_{NaCl}=\frac{36.100%}{136+100}\approx 26,47%`

`C%_{\text{nước đường}}=\frac{204.100%}{204+100}\approx 67,1%`

`b)` `n_{KOH}=\frac{5,6}{39+17}=0,1(mol)`

`n_{NaCl}=\frac{5,85}{58,5}=0,1(mol)`

`V=150ml=0,15l`

`CM_{NaCl}=CM_{KOH}=\frac{n}{V}=\frac{0,1}{0,15}\approx 0,67M`