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MHT CET · Chemistry · Solutions

The molar concentration of chloride ions in the resulting solution of \(300 \mathrm{~mL}\) of \(3.0 \mathrm{M} \mathrm{NaCl}\) and \(200 \mathrm{~mL}\) of \(4.0 \mathrm{M} \mathrm{BaCl}_{2}\) will be

  1. A \(1.7 \mathrm{M}\)
  2. B \(1.8 \mathrm{M}\)
  3. C \(5.0 \mathrm{M}\)
  4. D \(3.5 \mathrm{M}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(5.0 \mathrm{M}\)

Step-by-step Solution

Detailed explanation

\(\underset{4.0 \mathrm{M}}{\mathrm{BaCl}_{2}} \rightleftharpoons \mathrm{Ba}^{2+}+\underset{2 \times 4.0 \mathrm{M}}{2 \mathrm{Cl}^{-}}\)
\(\therefore\) Molar concentration of
\(
\begin{aligned}
\mathrm{Cl}^{-} &=\frac{M_{1} V_{1}+M_{2} V_{2}}{V_{1}+V_{2}} \\
&=\frac{3.0 \times 300+2 \times 4.0 \times 200}{300+200} \\
&=\frac{900+1600}{500} \\
&=5.0 \mathrm{M}
\end{aligned}
\)