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

a) Ta có: \(C\%_{NaCl}=\dfrac{60}{60+1250}\cdot100\%\approx4,58\%\)

b) Ta có: \(C_{M_{Na_2CO_3}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

a.\(C\%_{NaCl}=\dfrac{9}{91+9}.100\%=9\%\)

b.\(m_{NaCl}=0,5.58,5=29,25g\)

\(C\%_{NaCl}=\dfrac{29,25}{29,25+300}.100\%=8,88\%\)

`C_[M_[NaCl]]=[0,2]/[0,5]=0,4(M)`

C_{mnaCl =} \(\dfrac{0,2}{0,5}=0,4\left(M\right)\)

a) \(C\%=\dfrac{9}{9+91}.100\%=9\%\)

b) \(m_{H_2O}=300.1=300\left(g\right)\)

\(C\%=\dfrac{0,5.58,5}{0,5.58,5+300}.100\%=8,88\%\)

\(n_{Na_2O}=\dfrac{6.2}{62}=0.1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(0.1...........................0.2\)

\(C_{M_{NaOH}}=\dfrac{0.2}{1}=0.2\left(M\right)\)

nCuSO4=40/160=0,25 mol

CM CuSO4 =0,25/0,1=2,5M

nNaCl = 30/58,5=20/39 mol

nH2O = 170 /18=85/9 mol

2NaCl + 2H2O -->  Cl2 +     H2      +           2NaOH

20/39                    10/39      10/39                 20/39              mol        

 ta thấy nNaCl/2<nH2O/2

=> NaCl hết , H2O dư

=>mNaOH=20/39*20\(\approx\)20,51 g

m dd sau = 30 + 170 - 10/39*35,5-10,39*2\(\approx\)190,38 g

C% NaOh = 20,51*100/190,38=10,77%

Ta có: \(n_{Na_2O}=\dfrac{18,6}{62}=0,3\left(mol\right)\)

PTHH: Na₂O + H₂O ---> 2NaOH

Theo PT: \(n_{NaOH}=2.n_{Na_2O}=2.0,3=0,6\left(mol\right)\)

=> \(C_{M_{NaOH}}=\dfrac{0,6}{0,4}=1,5M\)

250ml = 0,25 lít

\(C_{M_{KOH}}=\dfrac{0,5}{0,25}=2M\)

CM KOH=\(\dfrac{0,5}{0,25}\)=2M