Hòa tan 36,5 gam HCl vào nước, thu được 500ml dung dịch có khối lượng riêng D = 1,1 g/ml. Tính nồng độ mol và nồng độ phần trăm của dung dịch thu được.
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(C\%=\dfrac{30}{170}.100\%=17,647\%\)
\(V_{\text{dd}}=\left(30+170\right)1,1=220ml\)
\(n_{NaCl}=\dfrac{30}{58,5}=0,513mol\)
\(C_M=\dfrac{0,513}{0,22}=0,696M\)
\(C\%_{NaCl}=\dfrac{30}{170+30}.100\%=15\%\\ C_M=C\%.\dfrac{10D}{M}=10.\dfrac{10.1,1}{58,5}=1,88M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\)
\(\Rightarrow C_{MddHCl}=\dfrac{1}{0,5}=2M\)
\(m_{ddHCl}=500.1,1=550\left(g\right)\)
\(\Rightarrow C\%ddHCl=\dfrac{36,5}{550}.100\%\approx6,6\%.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi $n_{Na_2O} = 2a(mol) \Rightarrow n_{K_2O} = a(mol)$
$\Rightarrow 2a.62 + 94a = 21,8 \Rightarrow a = 0,1(mol)$
$Na_2O + H_2O \to 2NaOH$
$K_2O + H_2O \to 2KOH$
$n_{NaOH} = 2n_{Na_2O} = 0,4(mol)$
$n_{KOH} = 2n_{K_2O} = 0,2(mol)$
$C_{M_{NaOH}} = \dfrac{0,4}{0,5} = 0,8M$
$C_{M_{KOH}} = \dfrac{0,2}{0,5} = 0,4M$
$m_{dd} = D.V = 1,04.500 = 520(gam)$
$C\%_{NaOH} = \dfrac{0,4.40}{520}.100\% = 3,1\%$
$C\%_{KOH} = \dfrac{0,2.56}{520}.100\% = 2,15\%$
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: 9,2 gam Na
\(a,n_{Na_2O}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
0,4------------------>0,8
\(\rightarrow C_{M\left(NaOH\right)}=\dfrac{0,8}{0,5}=1,6M\)
\(b,n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)
PTHH: \(K_2O+H_2O\rightarrow2KOH\)
0,4----------------->0,8
\(\rightarrow C\%_{KOH}=\dfrac{0,8.56}{362,4+37,6}.100\%=11,2\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{P_2O_5}=\dfrac{14.2}{142}=0.1\left(mol\right)\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
\(0.1............................0.2\)
\(m_{H_3PO_4}=0.2\cdot98=19.6\left(g\right)\)
\(m_{dd_{H_3PO_4}}=14.2+185.8=200\left(g\right)\)
\(C\%H_3PO_4=\dfrac{19.6}{200}\cdot100\%=9.8\%\)
\(V_{dd_{H_3PO_4}}=\dfrac{200}{1.1}=181.8\left(ml\right)=0.1818\left(l\right)\)
\(C_{M_{H_3PO_4}}=\dfrac{0.2}{0.1818}=1.1\left(M\right)\)
Ta có: \(n_{P_2O_5}=\dfrac{14,2}{142}=0,1\left(mol\right)\)
PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
____0,1_____________0,2 (mol)
\(\Rightarrow m_{H_3PO_4}=0,2.98=19,6\left(g\right)\)
Có: m dd sau pư = mP2O5 + mH2O = 200 (g)
\(\Rightarrow C\%_{H_3PO_4}=\dfrac{19,6}{200}.100\%=9,8\%\)
Có: V dd sau pư = \(\dfrac{200}{1,1}=\dfrac{2000}{11}\left(ml\right)=\dfrac{2}{11}\left(l\right)\)
\(\Rightarrow C_{M_{H_3PO_4}}=\dfrac{0,2}{\dfrac{2}{11}}=1,1M\)
Bạn tham khảo nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Na+H_2O->NaOH+\dfrac{1}{2}H_2\\ a.n_{Na}=\dfrac{m_1}{23}\left(mol\right)\\ m_{ddsau}=\dfrac{m_1}{23}+m_2-\dfrac{m_1}{46}=\dfrac{m_1}{46}+m_2\left(g\right)\\ C\%_B=\dfrac{\dfrac{40}{23}m_1}{\dfrac{m_1}{46}+m_2}\cdot100\%.\\ b.C_M=\dfrac{10dC\%}{M}=10\cdot1,2\cdot\dfrac{0,05}{40}=0,015\left(M\right)\)
\(Na+H_2O->NaOH+\dfrac{1}{2}H_2\\ a.n_{Na}=\dfrac{m_1}{23}\left(mol\right)\\ m_{ddsau}=m_1+m_2-\dfrac{m_1}{23}=\dfrac{22}{23}m_1+m_2\left(g\right)\\ C\%_B=\dfrac{\dfrac{40}{23}m_1}{\dfrac{22}{23}m_1+m_2}\cdot100\%.\\ b.C_M=\dfrac{10dC\%}{M}=10\cdot1,2\cdot\dfrac{0,05}{40}=0,015\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(n_{FeSO_4.7H_2O}=\dfrac{41,7}{278}=0,15\left(mol\right)\)
=> \(n_{FeSO_4}=0,15\left(mol\right)\)
=> \(m_{FeSO_4}=0,15.152=22,8\left(g\right)\)
b) mdd sau pha trộn = 41,7 + 207 = 248,7 (g)
c) \(C\%=\dfrac{22,8}{248,7}.100\%=9,168\%\)
\(V_{dd}=\dfrac{248,7}{1,023}=243,1085\left(ml\right)=0,2431085\left(l\right)\)
\(C_M=\dfrac{0,15}{0,2431085}=0,617M\)