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:

\(C\%=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{10}{50}.100\%=20\%\)

\(a.\)

\(m_{dd}=10+40=50\left(g\right)\)

\(C\%=\dfrac{10}{50}\cdot100\%=20\%\)

\(b.\)

\(m_{KOH}=0.25\cdot56=14\left(g\right)\)

\(m_{dd_{KOH}}=14+36=50\left(g\right)\)

\(C\%_{KOH}=\dfrac{14}{50}\cdot100\%=28\%\)

\(m_{ddKNO_3}=40+3600=3640g\\ C_{\%KNO_3}=\dfrac{40}{3640}\cdot100\%=1,1\%\)

_ _ _ _ _ _ _

\(3600g\rightarrow360g\\ m_{ddKNO_3}=40+360=400g\\ C_{\%KNO_3}=\dfrac{40}{400}\cdot100\%=10\%\)

10%

10%

10%

\(a,b,m_{dd}=40+200=240\left(g\right)\\ C\%_{C_{12}H_{22}O_{11}}=\dfrac{40}{240}.100\%=16,67\%\)

\(a,C\%_{NaCl}=\dfrac{10}{10+90}.100\%=10\%\\ b,C\%_{NaCl}=\dfrac{10}{10+90+150}.100\%=6,67\%\)

10,00%

Bài 2

\(C_{\%đường}=\dfrac{10}{10+100}\cdot100\%\approx9,09\%\)

Bài 3

\(n_{NaOH}=\dfrac{4}{40}=0,1mol\\ C_{M_{NaOH}}=\dfrac{0,1}{0,2}=0,5M\)

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%

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: n_{NaOH (thêm vào)} = a (g)

\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)