`m_(NaOH)=12+((78.16)/100)=24,48g`

`%C_(NaOH)=((24,48)/(12+78)).100%=27,2%`

B) m dd =500g

m HCl= (200.12%)+(300.18%)=78g

C%= 78/500 .100%= 15,6%

Suy ra :

$\dfrac{150}{460} = \dfrac{6\% -x\%}{4\%} \Rightarrow x = 0,047$

Ta có : 

\(m_{dd\ sau\ pư} = M_1 +M_2(gam)\\ \Rightarrow M_1.15\% + M_2.30\% = (M_1 + M_2).20\%\\ \Leftrightarrow 0,15M_1 + 0,3M_2 = 0,2M_1 + 0,2M_2\\ \Leftrightarrow 0,05M_1 = 0,1M_2\\ \Leftrightarrow \dfrac{M_1}{M_2} = \dfrac{0,1}{0,05} = 2\)

\(m_{NaOH}=10+150.20\%=40\left(g\right)\\ C\%_{NaOH}=\dfrac{40}{150+10}.100\%=25\%\)

$SO_3 + H_2O \to H_2SO_4$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = n_{SO_3} = \dfrac{12}{80} = 0,15(mol)$
$n_{NaOH} = 2n_{H_2SO_4} = 0,3(mol)$

$m_{NaOH} = 0,3.40 = 12(gam)$

a) \(m_{NaOH}=\dfrac{60.20}{100}=12\left(g\right)\)

\(C\%_{dd.sau.khi.pha}=\dfrac{12}{60+40}.100\%=12\%\)

b) \(C\%_{dd.sau.khi.pha}=\dfrac{12+12}{60+12}.100\%=33,33\%\)

60g dd 20% có 60.20%=12g NaOH, 60-12=48g H2O

a, Pha thêm 40g H2O , ta có 88g H2O

→C%NaOH=12.100:88=12,64% 

b) 

tan thêm 12 g

=>m NaOH=24g

=>C%=\(\dfrac{24}{60+12}100\)=33,33%

\(m_{NaOH}=\dfrac{60.20}{100}=12\left(g\right)\\ a,m_{\text{dd}}=60+40=100\left(g\right)\\ C\%=\dfrac{12}{100}.100\%=12\%\\ b,m_{NaOH}=12+12=24\left(g\right)\\ C\%=\dfrac{24}{60}.100\%=40\%\)

\(m_{ddCuSO4}=1,12.100=112\left(g\right)\)

\(m_{ct}=\dfrac{10.112}{100}=11,2\left(g\right)\)

\(n_{CuSO4}=\dfrac{11,2}{160}=0,07\left(mol\right)\)

Pt : \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4|\)

            1              2                   1                  1

          0,07        0,14

\(n_{NaOH}=\dfrac{0,07.2}{1}=0,14\left(mol\right)\)

\(V_{ddNaOH}=\dfrac{0,14}{2}=0,07\left(l\right)\)

 Chúc bạn học tốt

m dd=112 g

m CuSO4=11,2g =>n CuSO4=11,2\160=0,07 mol

CuSO4+2NaOH->Cu(OH)2+Na2SO4

0,07-------0,14 mol

=>V NaOH=0,14\2=0,07l