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Question

In an aqueous solution of barium nitrate, the [NO3] is 0.080 M. This solution can be labeled as

A
0.040 N(Ba(NO2)2
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B
0.160 MBa(NO3)2
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C
0.080 NBa(NO3)2
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D
0.080 MNO3
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Solution

The correct option is C 0.080 NBa(NO3)2
InBa(NO3)2+ the molar ratio of Ba(NO3)2 to NO3 is 1 : 2
Therefore, the molarity of the Ba(NO3)2 solution is
12×0.080=0.040 M
As n - factor of Ba(NO3)2 is 2, its normality will be 0.080 N as normality= molarity × valency factor.
So, the correct answer will be option (c).

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