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![](https://rs.olm.vn/images/avt/0.png?1311)
\(m_{dung\ dịch} =m_{NaCl} + m_{H_2O} = 45 + 155 = 200(gam)\\ C\%_{NaCl} = \dfrac{m_{NaCl}}{m_{dung\ dịch}}.100\% = \dfrac{45}{200}.100\% = 22,5\%\)
\(C\%=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{m_{ct}}{45+155}.100\%=\dfrac{45}{200}.100\%=22,5\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C\%_{KCl}=\dfrac{20}{600}\cdot100\%=3.33\%\)
\(C\%_{K_2SO_4}=\dfrac{75}{1500}\cdot100\%=5\%\)
\(C\%_{NaCl}=\dfrac{15}{15+45}\cdot100\%=25\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
a) \(C\%=\dfrac{15}{15+45}.100\%=25\%\)
b) \(C_M=\dfrac{0,5}{1,5}=0,33M\)
Câu 2:
a) \(n_{NaOH}=0,5.1=0,5\left(mol\right)=>m_{NaOH}=0,5.40=20\left(g\right)\)
b) \(n_{HCl}=0,2.0,5=0,1\left(mol\right)=>m_{HCl}=0,1.36,5=3,65\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : mdd = mct + mdm
\(\Rightarrow\)mdd =15 + 45 = 60 ( gam )
Áp dụng công thức \(C\%=\dfrac{m_{ct}}{m_{dd}}.100\%\) ta có :
\(C\%=\dfrac{15}{60}.100\%=25\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(C\%_{dd_{HCl}}=\dfrac{m_{HCl}}{mdd_{HCl}}\cdot100\%=\dfrac{20}{600}\cdot100\%=3,33\%\)
b)
\(mdd_{NaCl}=m_{NaCl}+m_{H_2O}=15+45=60\left(g\right)\)
=> \(C\%dd_{NaCl}=\dfrac{m_{NaCl}}{mdd_{NaCl}}\cdot100\%=\dfrac{15}{60}\cdot100\%=25\%\)
hok tốt nhé
\(C\%_{NaCl}=\dfrac{15}{15+45}.100\%=25\%\)
\(∑m_{dd}= m_{NaCl} + m_{H_2O} = 15 + 45 = 60(gam)\)
\(\text{C%}ddNaCl=\)\(\dfrac{15}{60}.100\%=25\%\)