Question

The molar concentration of chloride ions in the resulting solution of $300mL$ of $3.0MNaCl$ and $200mL$ of $4.0MBaC{l}_2$ will be :

• A. $1.7M$
• B. $1.8M$
• C. $5.0M$
• D. $3.5M$

Answer 1

Answer: (C)

$5.0M$

$\underset{3.0M}{NaCl}\leftrightharpoons N{a}^{+}+\underset{3.0M}{C{l}^{-}}$

$\underset{4.0M}{BaC{l}_2}\leftrightharpoons B{a}^{2+}+\underset{2\times 4.0M}{2C{l}^{-}}$

$\therefore$ Molar concentration of

$C{l}^{-}=\frac{M_1{V}_1+M_2{V}_2}{V_1+V_2}$

$=\frac{3.0\times 300+2\times 4.0\times 200}{300+200}$

$=\frac{900+1600}{500}$

$=5.0M$.

Hence option C is correct.