\(n_{NaCl}=\dfrac{1,17}{58,5}=0,02\left(mol\right)\\ n_{BaCl_2}=\dfrac{2,08}{208}=0,01\left(mol\right)\\ \left[Na^+\right]=\dfrac{0,02}{0,1}=0,2\left(M\right)\\ \left[Ba^{2+}\right]=\dfrac{0,01}{0,1}=0,1\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,02+0,01.2}{0,1}=0,4\left(M\right)\)

a) 200ml=0,2(l)

Ta có: \(n_{KOH}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{n}{V}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)

b) 1500ml=1,5(l)

\(C_{M_{Na_2CO_3}}=\dfrac{n}{V}=\dfrac{0,06}{1,5}=0,04\left(M\right)\)

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

\(a.500ml=0,5l\\ C_{M_{Na_2CO_3}}=\dfrac{0,1}{0,5}=0,2M\\ b.m_{ddNa_2CO_3}=500.1=500g\\ m_{Na_2CO_3}=0,1.106=10,6g\\ C_{\%Na_2CO_3}=\dfrac{10,6}{500}\cdot100\%=2,12\%\)

\(a.\)

\(m_{NaCl}=20\left(g\right)\)

\(m_{dd_{NaCl}}=20+60=80\left(g\right)\)

\(C\%_{NaCl}=\dfrac{20}{80}\cdot100\%=25\%\)

\(b.\)

\(m_{K_2O}=9.4\left(g\right)\)

\(\Rightarrow n_{K_2O}=\dfrac{9.4}{94}=0.1\left(mol\right)\)

\(K_2O+H_2O\rightarrow2KOH\)

\(0.1.....................0.2\)

\(m_{KOH}=0.2\cdot56=11.2\left(g\right)\)

\(m_{dd_{KOH}}=9.4+90.6=100\left(g\right)\)

\(C\%_{KOH}=\dfrac{11.2}{100}\cdot100\%=11.2\%\)

a, Khối lượng chất tan : 20g

Khối lượng dung dịch : 20 + 60 = 80g

Nồng độ : C% = 25%

b, K2O + H2O -> 2KOH

..0,1.......................0,2....

- Khối lượng chất tan = mKOH = 11,2g

Khối lượng dung dịch = 100g

- Nộng độ : C%KOH = 11,2%

a)

\(m_{chất\ tan} = m_{NaCl} = 20(gam)\\ m_{dd} = 20 + 60 = 80(gam)\\ C\%_{NaCl} = \dfrac{20}{80}.100\%= 25\%\)

b)

\(n_{K_2O} = \dfrac{9,4}{94} = 0,1(mol)\\ K_2O + H_2O \to 2KOH\\ n_{KOH} = 2n_{K_2O} = 0,2(mol)\\ m_{chất\ tan} = m_{KOH} = 0,2.56 = 11,2(gam)\\ m_{dd} = 9,4 + 90,6 = 100(gam)\\ C\%_{KOH} = \dfrac{11,2}{100}.100\% = 11,2\%\)

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

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

_____________________________________________

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

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

_____________________________________________

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

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

_____________________________________________

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

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

\(n_{Na_2CO_3}=\dfrac{2,12}{106}=0,02\left(mol\right)\\ C_{MddNa_2CO_3}=\dfrac{0,02}{2}=0,01\left(M\right)\)

\(n_{K_2SO_4}=\dfrac{1,74}{174}=0,01\left(mol\right)\)

\(\Rightarrow C_{M\left(K_2SO_4\right)}=\dfrac{0,01}{0,4}=0,025M\)

Phương trình điện li: \(K_2SO_4\rightarrow2K^++SO_4^{2-}\)

\(\Rightarrow\left[K^+\right]=2C_{M\left(K_2SO_4\right)}=0,05M\)

\(\left[SO_4^{2+}\right]=C_{M\left(K_2SO_4\right)}=0,025M\)

lm ơn giúp mình nha