The molar concentration of the chloride ion in the solution obtained by mixing $300 m L$ of $3.0 M N a C l$ and $200 m L$ of $4.0 M$ solution of $B a C l_{2}$ is:

1.6 M

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1.8 M

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5.0 M

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0.5 M

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Solution

The correct option is C $5.0 M$
By applying normality formula ,
$\frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}} = M_{3}$
$∵ N_{a} C l n f a c t o r = 1$
$B_{a} C l_{2} n f a c t o r = 2$
$∴ \frac{1 \times 300 \times 3 + 200 \times 4 \times 2}{300 + 200} = M_{3}$
$\Rightarrow \frac{900 + 1600}{500} = M_{3}$
Molarity of Chloride ion = $\frac{2500}{500} = 5.0 M$

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