If 50 ml of 0.1 M $N a C l$ and 50 ml of 0.1 M $B a C l_{2}$ are mixed, molarity of chloride ion in the resulting solution will be:

0.2 M

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0.3 M

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0.15 M

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0.1 M

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Solution

The correct option is C $0.15 M$
50 ml of 0.1 M $N a C l$ we have 0.005 moles of chloride ions.

In 50ml of 0.1 M $B a C l_{2}$ we have 0.01 moles of chloride ions

So, total moles of chloride ions $= 0.01 + 0.005 = 0.015$

Total volume of solution $= 100 m l$

So, molarity of chloride ion = $\frac{0.015}{100} \times 1000 = 0.15 M$
Hence, the correct option is C.

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