1)

$m_{dd} = 50 + 30 = 80(gam)$
$m_{KOH} = 50.20\% + 30.15\% = 14,5(gam)$
$C\% = \dfrac{14,5}{80}.100\% = 18,125\%$

2)

$m_{dd} = 200 + 300 = 500(gam)$
$m_{NaCl} = 200.20\% + 300.5\% = 55(gam)$

$C\% = \dfrac{55}{500}.100\% = 11\%$

3)

4)

$V_{dd} = 0,3 + 0,2 = 0,5(lít)$
$n_{H_2SO_4} = 0,3.1,5 + 0,2.2 = 0,85(mol)$
$C_{M_{H_2SO_4}} = \dfrac{0,85}{0,5} = 1,7M$

m₁ = \(\dfrac{80\times10}{100}=8\left(g\right)\)

m₂ = \(\dfrac{20\times20}{100}=4\left(g\right)\)

C% dd thu được = \(\dfrac{8+4}{80+20}.100\%=12\%\)

2

b

mNaCl=\(\dfrac{200.15}{100}\)=30(g)

nNaCl=\(\dfrac{30}{58,5}\)=0.51(mol)

VddNaCl=\(\dfrac{200}{1,1}\)=181.8(ml)=0.1818(l)

CMNaCl=\(\dfrac{0,51}{0,1818}\)=2.8(M)

a) \(m_{NaCl}=80\times15\%=12\left(g\right)\)

\(m_{ddNaCl}mới=20+80=100\left(g\right)\)

\(\Rightarrow C\%_{NaCl}mới=\frac{12}{100}\times100\%=12\%\)

b) \(m_{NaCl.20\%}=200\times20\%=40\left(g\right)\)

\(m_{NaCl.5\%}=300\times5\%=15\left(g\right)\)

\(\Rightarrow m_{NaCl}mới=40+15=55\left(g\right)\)

\(m_{ddNaCl}mới=200+300=500\left(g\right)\)

\(\Rightarrow C\%_{NaCl}mới=\frac{55}{500}\times100\%=11\%\)

c) \(m_{H_2SO_4.10\%}=100\times10\%=10\left(g\right)\)

\(m_{H_2SO_4.25\%}=150\times25\%=37,5\left(g\right)\)

\(\Rightarrow m_{H_2SO_4}mới=10+37,5=47,5\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}mới=100+150=250\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}mới=\frac{47,5}{250}\times100\%=19\%\)

Áp dụng quy tắc đường chéo ta có:

a) \(D_1=20g\)

\(D_2=80g\)

\(\frac{D_1}{D_2}=\frac{20}{80}=\frac{15-C\%}{C\%}\rightarrow C\%=12\%\)

b) \(D_1=200\left(g\right)\)

\(D_2=300\left(g\right)\)

\(\frac{D_1}{D_2}=\frac{200}{300}=\frac{C\%-5}{20-C\%}\rightarrow C\%=11\%\)

c) \(D_1=100\left(g\right)\)

\(D_2=150\left(g\right)\)

\(\frac{D_1}{D_2}=\frac{100}{150}=\frac{25-C\%}{C\%-10}\rightarrow C\%=19\%\)

Tính nồng độ phần trăm của dung dịch trong các trường hợp sau:

a) Cho 20 gam NaCl tan hoàn toàn trong 80 gam H2O?

-> mddNaCl= mNaCl+mH2O= 20+80=100(g)

=>C%ddNaCl= (mNaCl/mddNaCl).100%= (20/100).100= 20%

b) Cho 50 gam H2O vào 100 gam dung dịch HCl 10%?

mHCl= 10%.100= 10(g)

=> mddHCl(sau)= mddHCl(10%)+mH2O= 100+50=150(g)

=>C%ddHCl(sau)= (mHCl/mddHCl(sau)).100%= (10/150).100\(\approx6,667\%\)

c) Trộn 200 gam dung dịch HCl 20% với 100 gam dung dịch HCl 10%?

--

mHCl(tổng)= 200.20%+100.10%= 50(g)

mddHCl(tổng)=200+100=300(g)

=> C%ddHCl(tổng)= (mHCl(tổng)/ mddHCl(tổng)).100%= (50/300).100\(\approx16,667\%\)

d) Cho thêm 10 gam NaCl vào 90 gam dung dịch NaCl 30%?

---

mNaCl(trong dd 30%)= 30%.90=27(g)

=>mNaCl(tổng)= 10+27=37(g)

mddNaCl(tổng)=10+90=100(g)

=>C%ddNaCl(sau)= (mNaCl(tổng)/ mddNaCl(tổng)).100%= (37.100)/100= 37%

\(n_{NaCl}=\dfrac{11,7}{58,5}=0,2\left(mol\right)\)

\(n_{NaNO_3}=\dfrac{100.8,5\%}{85}=0,1\left(mol\right)\)

\(V_{dd}=\dfrac{100}{1,25}=80\left(ml\right)\)

\(\left\{{}\begin{matrix}C_{M\left(NaCl\right)}=\dfrac{0,2}{0,08}=2,5M\\C_{M\left(NaNO_3\right)}=\dfrac{0,1}{0,08}=1,25M\end{matrix}\right.\)

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