\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\\ C_{M_{NaOH}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)

n NaOH= \(\dfrac{4}{40}\)=0,1(mol)

C MnaoH=\(\dfrac{0,1}{0,4}\)=0,25(M)

\(1,C_{M\left(HCl\right)}=\dfrac{0,75}{0,5}=1,5M\\ 2,n_{Ca\left(OH\right)_2}=\dfrac{37}{74}=0,5\left(mol\right)\\ C_{M\left(Ca\left(OH\right)_2\right)}=\dfrac{0,5}{1,5}=0,33M\\ 3,n_{NaOH}=0,25+\dfrac{20}{40}=0,75\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,75}{2}=0,375M\\ 4,n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,5}{2}=0,25M\)

`1) C_[M_[HCl]] = [ 0,75 ] / [ 0,5 ] = 1,5 (M)`

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`2)n_[Ca(OH)_2] = 37 / 74 = 0,5 (mol)`

`-> C_[M_[Ca(OH)_2]] = [ 0,5 ] / [ 1,5 ] ~~ 0,33 (M)`

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`3) n_[NaOH] = 0,25 + 20 / 40 = 0,75 (mol)`

`-> C_[M_[NaOH]] = [ 0,75 ] / 2 = 0,375 (M)`

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`4) n_[H_2 SO_4] = 49 / 98 = 0,5 (mol)`

`-> C_[M_[H_2 SO_4]] = [ 0,5 ] / 2 = 0,25 (M)`

Bài 2

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

Bài 3

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

\(a,n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\\ C_{M\left(HCl\right)}=\dfrac{0,2}{0,4}=0,5M\\ b,n_{H_2SO_4}=\dfrac{73,5}{98}=0,75\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,75}{0,5}=1,5M\\ n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,4}{0,25}=1,6M\\ n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\\ C_{M\left(Ba\left(OH\right)_2\right)}=\dfrac{0,2}{0,8}=0,25M\)

Ta có: \(n_{Na_2CO_3}=n_{Na_2CO_3.10H_2O}=\dfrac{38,61}{286}=0,135\left(mol\right)\)

m _{dd sau pư} = 38,61 + 256 = 294,61 (g)

\(\Rightarrow C\%_{Na_2CO_3}=\dfrac{0,135.106}{294,61}.100\%\approx4,86\%\)

Có: \(V_{ddsaupư}=\dfrac{294,61}{1,156}\approx254,85\left(ml\right)\approx0,255\left(l\right)\)

\(\Rightarrow C_{M_{Na_2CO_3}}=\dfrac{0,135}{0,255}\approx0,53M\)

Bạn tham khảo nhé!

Gọi số mol của Na₂CO₃ là a (mol) \(\Rightarrow n_{H_2O\left(phân.tử\right)}=10a\left(mol\right)\)

\(\Rightarrow106a+18\cdot10a=38,61\) \(\Leftrightarrow a=0,135\left(mol\right)\)

\(\Rightarrow C\%_{Na_2CO_3}=\dfrac{0,135\cdot106}{38,61+256}\cdot100\%\approx4,86\%\)

Mặt khác: \(V_{ddNa_2CO_3}=\dfrac{38,61+256}{1,156}\approx254,41\left(ml\right)\) \(\Rightarrow C_{M_{Na_2CO_3}}=\dfrac{0,135}{0,25441}\approx0,53\left(M\right)\)

a)\(n_{K_2O}=\dfrac{23,5}{94}=0,25mol\)

\(K_2O+H_2O\rightarrow2KOH\)

0,25      0,25       0,5

\(C_M=\dfrac{0,5}{0,5}=1M\)

b)Để trung hòa: \(n_{H^+}=n_{OH^-}=0,5\)

   \(\Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{H^+}=0,25mol\)

   \(m_{H_2SO_4}=0,25\cdot98=24,5g\)

   \(\Rightarrow m_{ddHCl}=\dfrac{24,5\cdot100\%}{60\%}=\dfrac{245}{6}g\)

   Thể tích dung dịch:

   \(V=\dfrac{m}{D}=\dfrac{\dfrac{245}{6}}{1,5}\approx27,22ml\)

\(n_{K_2O}=\dfrac{23,5}{94}=0,25\left(mol\right)\\ K_2O+H_2O\rightarrow2KOH\\ n_{KOH}=2.0,25=0,5\left(mol\right)\\ a,C_{M\text{dd}A}=C_{M\text{dd}KOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b,2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ n_{H_2SO_4}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{H_2SO_4}=0,25.98=24,5\left(g\right)\\ m_{\text{dd}H_2SO_4}=\dfrac{24,5.100}{60}=\dfrac{245}{6}\left(g\right)\\ V_{\text{dd}H_2SO_4}=\dfrac{\dfrac{245}{6}}{1,5}=\dfrac{245}{9}\left(ml\right)\approx27,222\left(ml\right)\)

Đáp án A

\(n_{KOH}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(C_{M_{KOH}}=\dfrac{0.1}{0.4}=0.25\left(M\right)\)