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Question

0.45 g of an acid (molar mass: 90 g/mol) was neutralized by 20 mL of a 0.5 N caustic potash. The basicity of the acid is:


A
1
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B
3
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C
2
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D
4
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Solution

The correct option is B 2
Let the basicity (n-factor) of the acid be x.
Equivalents of the acid = equivalents ofKOH (caustic potash) 
x×0.4590=0.51000×20
x=0.51000×20×900.45
On solving, x=2
So, the basicity is 2.

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