\(a,C_{M\left(NaOH\right)}=\dfrac{0,3}{0,5}=0,6M\\ b,n_{NaOH}=\dfrac{24}{40}=0,6\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,6}{0,4}=1,5M\)

a, \(C_{M_{H_2SO_4}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\)

b, \(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,4}=1,25\left(M\right)\)

\(a)C_{M\left(H_2SO_4\right)}=\dfrac{0,1}{0,25}=0,4M\\ b)C_{M\left(NaOH\right)}=\dfrac{\dfrac{20}{40}}{0,4}=1,25M\)

\(a,n_{H_2SO_4}=0,1mol\)

\(V_{H_2SO_4}=0,25l\left(250ml=0,25l\right)\\ C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,1}{0,25}=0,4M\)

`b,` \(n_{NaOH}=\dfrac{m}{M}=\dfrac{20}{40}=0,5mol\)

\(V_{NaOH}=0,4l\left(400ml=0,4l\right)\)

\(C_{M_{NaOH}}=\dfrac{n}{V}=\dfrac{0,5}{0,4}=1,25M\)

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) \(n_{HCl}=0,2.0,5=0,1\left(mol\right)=>m_{HCl}=0,1.36,5=3,65\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,1}{0,25}=0,4M\)

Bài 1:

\(n_{KNO_3}=\dfrac{20}{101}=0,198\left(mol\right)\)

\(C_M=\dfrac{n}{V}=\dfrac{0,198}{0,85}=0,233M\)

Bài 2:

\(C_M=\dfrac{n}{V}=\dfrac{0,5}{0,75}=0,66M\)

Bài 3:

\(n_{KNO_3}=2.0,5=1\left(mol\right)\)

\(m_{KNO_3}=1.101=101\left(g\right)\)

Bài 4:

\(C\%=\dfrac{20}{600}.100=3,33\%\)

Bài 1:

\(n_{KNO_3}=\dfrac{20}{101}=0,198\left(mol\right)\)

\(C_{M_{ddKNO_3}}=\dfrac{0,198}{0,85}\approx0,23M\)

Bài 2:

\(C_{M_{ddKCl}}=\dfrac{0,5}{0,75}\approx0,667M\)

Bài 3:

\(n_{KNO_3}=0,5.2=1\left(mol\right)\Rightarrow m_{KNO_3}=1.101=101\left(g\right)\)

Bài 4:

\(C\%_{ddKCl}=\dfrac{20.100\%}{600}=3,333\%\)

a) mNaOH= 200.10%=20(g)

b) nNaOH=0,4(mol)

=>CMddNaOH=0,4/0,2=2(M)

a) 

mNaOH = 200*10/100 = 20 (g) 

b) 

nNaOH = 16/40 = 0.4 (mol) 

CM NaOH = 0.4/0.2 = 2 (M) 

1. Dung dịch CuSO₄ có nồng độ 0,5mol/lit là trong 1 lít dung dịch CuSO₄ thì chứa 0,5 mol CuSO₄

2. Dung  dịch đường có nồng độ 2 mol/lit cho biết trong 1 lít dung dịch đường thì chứa 2 mol đường

3. \(CM_{NaCl}=\dfrac{n}{V}=\dfrac{0,75}{4}=0,1875M\)

4. \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\\ CM_{NaOH}=\dfrac{0,4}{0,2}=2M\)