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![](https://rs.olm.vn/images/avt/0.png?1311)
Khối lượng dung dịch là:
\(m_{dd}=m_{ct}+m_{dm}=10+40=50\left(g\right)\)
Nồng độ phần trăm của dung dịch là:
\(C\%=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{10}{50}.100\%=20\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\)
\(m_{dd}=10+40=50\left(g\right)\)
\(C\%=\dfrac{10}{50}\cdot100\%=20\%\)
\(b.\)
\(m_{KOH}=0.25\cdot56=14\left(g\right)\)
\(m_{dd_{KOH}}=14+36=50\left(g\right)\)
\(C\%_{KOH}=\dfrac{14}{50}\cdot100\%=28\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,b,m_{dd}=40+200=240\left(g\right)\\ C\%_{C_{12}H_{22}O_{11}}=\dfrac{40}{240}.100\%=16,67\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,C\%_{NaCl}=\dfrac{10}{10+90}.100\%=10\%\\ b,C\%_{NaCl}=\dfrac{10}{10+90+150}.100\%=6,67\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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%
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(C\%_{KCl}=\dfrac{20}{20+60}.100\%=25\%\)
b, \(C\%=\dfrac{40}{40+150}.100\%\approx21,05\%\)
c, \(C\%_{NaOH}=\dfrac{60}{60+240}.100\%=20\%\)
d, \(C\%_{NaNO_3}=\dfrac{30}{30+90}.100\%=25\%\)
e, \(m_{NaCl}=150.60\%=90\left(g\right)\)
f, \(m_{ddA}=\dfrac{25}{10\%}=250\left(g\right)\)
g, \(n_{NaOH}=120.20\%=24\left(g\right)\)
Gọi: nNaOH (thêm vào) = a (g)
\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)
\(C\%=\dfrac{10}{10+40}.100\%=20\%\)