$n_{NaCl(1)}$ = $C_{M(1)}.V_{NaCl(1)}=0,2.3=0,6(mol)$
$m_{NaCl(1)}$ = $n_{NaCl(1)}$ . $M_{NaCl}=0,6.58,5=35,1(g)$
$n_{NaCl(2)}$ = $C_{M(2)}.V_{NaCl(2)}=0,3.4=1,2(mol)$
$m_{NaCl(2)}$ = $n_{NaCl(2)}$ . $M_{NaCl}=1,2.58,5=70,2(g)$
Áp dụng phương pháp đường chéo
35,1 g dung dịch $NaCl_{}$0,2M $0,3-C_{M}$
$C_{M}$
70,2 g dung dịch $NaCl_{}$0,3M $ C_{M}-0,2$
⇒$\frac{35,1}{70,2}$ = $\frac{0,3-C_{M}}{C_{M}-0,2}$
⇒$35,1.(C_{M}-0,2_{})=70,2.(0,3-C_M)$
⇒$35,1C_{M}-7,02=21,06-70,2C_M$
⇒$35,1C_{M}+70,2C_M$=21,06+7,02
⇒$105,3C_{M}=28,08$
⇒$C_{M}$= $\frac{28,08}{105,3}≈0,27(M)$
