Đáp án C

n N a 2 S O 4 = 0,02 mol; n_NaCl=0,5 mol; n_Na+= 0,02.2+0,5= 0,54 mol

[Na⁺]= 0,54/(0,117+0,171+0,212)= 1,08M

n NaCl = 11,7/58,5 = 0,2(mol)

=> CM NaCl = 0,2/0,5 = 0,4M

500ml=0,5l

n_NaCl là 11,7/58,5=0,2 mol

=> C_M  = 0,2/0,5=0,4M

nNaCl=0,2*1,5=0,3mol

Sau khi thêm nước thì nNaCl không đổi

Vdd mới= 0,2+0,3=0,5(l)

suy ra Cm NaCl=0,3/0,5=0,6M

\(CM_{KCl}=\dfrac{n}{V}=\dfrac{1}{0,5}=2M\)

Đổi 500ml = 0,5l

\(C_{M\left(C\text{uS}O_4\right)}=\dfrac{n}{V_{\text{dd}}}=\dfrac{0,6}{0,5}=1,2M\)

\(C\%_{NaCl}=\dfrac{60}{400}\cdot100\%=15\%\)

Ta có: \(\left\{{}\begin{matrix}n_{NaCl}=\dfrac{60}{58,5}=\dfrac{40}{39}\left(mol\right)\\V_{NaCl}=\dfrac{400}{1,1}=\dfrac{4000}{11}\left(ml\right)=\dfrac{4}{11}\left(l\right)\end{matrix}\right.\) \(\Rightarrow C_{M_{NaCl}}=\dfrac{\dfrac{40}{39}}{\dfrac{4}{11}}\approx2,82\left(M\right)\)

Ta có: \(C\%_{NaCl}=\dfrac{60}{400}.100\%=15\%\)

\(V_{ddNaCl}=\dfrac{400}{1,1}=\dfrac{4000}{11}\left(ml\right)=\dfrac{4}{11}\left(l\right)\)

\(n_{NaCl}=\dfrac{60}{58,5}=\dfrac{40}{39}\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{\dfrac{40}{39}}{\dfrac{4}{11}}\approx2,82M\)

Bạn tham khảo nhé!