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Question

5.3 g of M2CO3 is dissolved in 150 mL of 1 N HCl, the unused acid required 100 mL of 0.5 N NaOH. Hence equivalent weight of M is:

A
23
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B
12
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C
24
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D
13
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Solution

The correct option is A 23
Milliequivalents of Acid used = Milliequivalents of base taken

Milliequivalent of acid used
=150×1100×0.5=100 meq
Equivalent of acid used = 0.1 equivalents
Equivalents of M2CO3 = Equivalents of Acid used = 0.1 equivalents

Equivalent weight of M2CO3 (by data) = Equivalent weight of M2CO3 (by formula)

given weightno. of equivalents= Molar weightn-factor=Equivalent weight

Let 'x' be the atomic weight of M
5.30.1=2x+602
x=23

Equivalent weight of M=231=23
as M is present as M+ in M2CO3, its n-factor is 1.

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