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
- A \(1.7 \mathrm{M}\)
- B \(1.8 \mathrm{M}\)
- C \(5.0 \mathrm{M}\)
- D \(3.5 \mathrm{M}\)
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
\(
\begin{aligned}
\mathrm{Cl}^{-} &=\frac{M_{1} V_{1}+M_{2} V_{2}}{V_{1}+V_{2}} \\
&=\frac{3.0 \times 300+2 \times 4.0 \times 200}{300+200} \\
&=\frac{900+1600}{500} \\
&=5.0 \mathrm{M}
\end{aligned}
\)
\(\therefore\) Molar concentration of
\(
\begin{aligned}
\mathrm{Cl}^{-} &=\frac{M_{1} V_{1}+M_{2} V_{2}}{V_{1}+V_{2}} \\
&=\frac{3.0 \times 300+2 \times 4.0 \times 200}{300+200} \\
&=\frac{900+1600}{500} \\
&=5.0 \mathrm{M}
\end{aligned}
\)
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