Đáp án B

\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\\ \rightarrow C_{M\left(Na_2CO_3\right)}=\dfrac{0,1}{0,2}=0,5M\)

Ta có: \(C\%=\dfrac{C_M.M}{10.D}\)

\(\rightarrow C\%=\dfrac{0,5.106}{10.1,05}=5,05\%\)

Ta có: \(m_{ddCuSO_4}=\dfrac{3}{15\%}=20\left(g\right)\)

\(V_{ddCuSO_4}=\dfrac{20}{1,15}\approx17,39\left(ml\right)\)

Ta có: \(n_{CuSO_4}=\dfrac{3}{160}=0,01875\left(mol\right)\)

\(\Rightarrow C_{M_{CuSO_4}}=\dfrac{0,01875}{0,01739}\approx1,08M\)

Bạn tham khảo nhé!

\(m_{dd_{HCl\left(10\%\right)}}=150\cdot1.206=180.9\left(g\right)\)

\(n_{HCl}=\dfrac{180.9\cdot10\%}{36.5}\approx0.5\left(mol\right)\)

\(n_{HCl\left(2M\right)}=0.25\cdot2=0.5\left(mol\right)\)

\(n_{HCl}=0.5+0.5=1\left(mol\right)\)

\(V_{dd_{HCl}}=150+250=400\left(ml\right)=0.4\left(l\right)\)

\(C_{M_{HCl}}=\dfrac{1}{0.4}=2.5\left(M\right)\)

1)

\(m_{ddCuSO_4\left(bd\right)}=1,6.25=40\left(g\right)\)

\(n_{CuSO_4.5H_2O}=\dfrac{11,25}{250}=0,045\left(mol\right)\)

=> \(n_{CuSO_4}=0,045\left(mol\right)\)

\(C_M=\dfrac{0,045}{0,025}=1,8M\)

\(C\%=\dfrac{0,045.160}{40}.100\%=18\%\)

b)

\(m_{CuSO_4}=\dfrac{200.18}{100}=36\left(g\right)\)

\(n_{CuSO_4.5H_2O}=\dfrac{5,634}{250}=0,022536\left(mol\right)\)

n_{CuSO4 (tách ra)} = 0,022536 (mol)

=> \(m_{CuSO_4\left(dd.ở.t^o\right)}=36-0,022536.160=32,39424\left(g\right)\)

\(m_{H_2O\left(bd\right)}=200-36=164\left(g\right)\)

n_{H2O (tách ra)} = 0,022536.5 = 0,11268 (mol)

=> \(m_{H_2O\left(dd.ở.t^o\right)}=164-0,11268.18=161,97176\left(g\right)\)

\(S_{t^oC}=\dfrac{32,39424}{161,97176}.100=20\left(g\right)\)

\(C\%=\dfrac{30}{170}.100\%=17,647\%\) 
\(V_{\text{dd}}=\left(30+170\right)1,1=220ml\) 
\(n_{NaCl}=\dfrac{30}{58,5}=0,513mol\)
\(C_M=\dfrac{0,513}{0,22}=0,696M\)

\(C\%_{NaCl}=\dfrac{30}{170+30}.100\%=15\%\\ C_M=C\%.\dfrac{10D}{M}=10.\dfrac{10.1,1}{58,5}=1,88M\)