a) \(C_{M_{MgCl_2}}=\frac{0,5}{0,75}=0,667\left(M\right)\)

b) \(n_{CuSO_4}=\frac{400}{160}=2,5\left(mol\right)\)

\(\Rightarrow C_{M_{CuSO_4}}=\frac{2,5}{4}=0,625\left(M\right)\)

c) \(C\%_{KCl}=\frac{20}{600}\times100\%=3,33\%\)

d) \(m_{ddNaCl}=20+180=200\left(g\right)\)

\(C\%_{NaCl}=\frac{20}{200}\times100\%=10\%\)

e) \(n_{KNO_3}=0,5\times2=1\left(mol\right)\)

\(\Rightarrow m_{KNO_3}=1\times101=101\left(g\right)\)

f) \(m_{MgCl_2}=50\times4\%=2\left(g\right)\)

\(n_{MgCl_2}=\frac{2}{95}\left(mol\right)\)

\(n_{CuSO_4}=\dfrac{20}{160}=0,125(mol)\\ a,CuSO_4+2NaOH\to Cu(OH)_2\downarrow+2NaCl\\ \Rightarrow n_{NaOH}=0,25(mol)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{0,25}{0,2}=1,25M\\ b,n_{Cu(OH)_2}=0,125(mol)\\ \Rightarrow m_{Cu(OH)_2}=0,125.98=12,25(g)\\ c,Cu(OH)_2\xrightarrow{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=0,125(mol)\\ \Rightarrow m_{CuO}=0,125.80=10(g)\)

a) Ta có: \(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,1}{0,5}=0,2\left(M\right)=\left[Na^+\right]=\left[Cl^-\right]\)

b) Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\) 

\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=0,4\left(M\right)\\\left[OH^-\right]=0,8\left(M\right)\end{matrix}\right.\)

c) Ta có: \(n_{H_2SO_4}=0,025\cdot2=0,05\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,05}{0,125+0,025}\approx0,33\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=0,66\left(M\right)\\\left[SO_4^{2-}\right]=0,33\left(M\right)\end{matrix}\right.\)

PTHH: 6NaOH + Al₂(SO₄)₃ --> 3Na₂SO₄ + 2Al(OH)₃

Số mol của Al₂(SO₄)₃ là: 0,04 : 6 = 1/150 mol

                  Đổi : 100mo = 0,1 lít

              C_M = 1/150 : 0,1 = 1/15 M

Đáp án B

ĐÁP ÁN B