Gọi : \(V_{NaOH(D = 1,4\ g/ml)} = a(ml) ; V_{NaOH(D= 1,1\ g/ml)} = b(ml)\)

Ta có : 

a + b = 600

Bảo toàn khối lượng : 1,4a + 1,1b = 600.1,2

Suy ra a = 200(ml) ; b = 400(ml)

a) Ta có: \(n_{Al\left(NO_3\right)_3}=\dfrac{4,26}{213}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al^+}=0,02\left(mol\right)\\n_{NO_3^-}=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[Al^+\right]=\dfrac{0,02}{0,1}=0,2\left(M\right)\\\left[NO_3^-\right]=\dfrac{0,06}{0,1}=0,6\left(M\right)\end{matrix}\right.\)

b) Ta có: \(\left[Na^+\right]=0,1+0,02\cdot2+0,3=0,304\left(M\right)\)

c) Bạn xem lại đề !!

\(a)2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\\ b)n_{CO_2}=\dfrac{1,1}{44}=0,025mol\\ n_{NaOH}=0,025.2=0,05mol\\ V_{ddNaOH}=\dfrac{0,05}{1,2}=\dfrac{1}{24}\approx0,04l\)

Tính nồng độ M của dung dịch sau:

a)Tính nồng độ M của dung dịch HCl 37%,D=1,19g/mol

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Gỉa sử có 100(g) dd HCl

-> mHCl= 100.37%= 37(g) => nHCl= 37/36,6 (mol)

VddHCl= mddHCl/DddHCl= 100/1,19 (ml)

=> \(C_{MddHCl}=\frac{n_{HCl}}{V_{ddHCl}}=\frac{\frac{37}{36,5}}{\frac{100}{1,19}}\approx0,012\left(M\right)\)

b)Tính nồng độ M của dung dịch KOH 12%,D=1,1g/mol

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Gỉa sử có 100(g) dd KOH

=> mKOH= 100.12%=12(g) => nKOH= 12/56(mol)

VddKOH= mddKOH / DddKOH= 100/1,1 (ml)

=> \(C_{MddKOH}=\frac{nKOH}{VddKOH}=\frac{\frac{12}{56}}{\frac{100}{1,1}}\approx0,002\left(M\right)\)

9% và 1,77M

Tong khoi luong chat ran la (30×5÷100)+(15×20:100)=4,5g

Tong khoi luong dd la 30+20=50g

C%= 4,5:50×100=9%

CM=n:V=mchattan/M.V=C%.mdd/100.M.V=C%.D.V.1000/100.M.V=1,77M

=>CM=C%.D.10/M=

=>CM=C%.D.10/M=

\(n_{P_2O_5}=\dfrac{14.2}{142}=0.1\left(mol\right)\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

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

\(m_{H_3PO_4}=0.2\cdot98=19.6\left(g\right)\)

\(m_{dd_{H_3PO_4}}=14.2+185.8=200\left(g\right)\)

\(C\%H_3PO_4=\dfrac{19.6}{200}\cdot100\%=9.8\%\)

\(V_{dd_{H_3PO_4}}=\dfrac{200}{1.1}=181.8\left(ml\right)=0.1818\left(l\right)\)

\(C_{M_{H_3PO_4}}=\dfrac{0.2}{0.1818}=1.1\left(M\right)\)

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} = m_P2O5 + m_H2O = 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!

`a)`

\(C_M=\dfrac{n}{V}=\dfrac{0,2}{4}=0,05\left(M\right)\)

`b)`

\(C_M=\dfrac{n}{V}=\dfrac{0,4}{8}=0,05\left(M\right)\)