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2.5 g of the carbonate of a metal war tr...

2.5 g of the carbonate of a metal war treated with 100 ml of `1 N H_(2)SO_(4)`. After the completion of the reaction, the solution was boiled off to expel `CO_(2)` and was then titrated against 1 N NaOH solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is 20.

A

50 ml

B

25 ml

C

75 ml

D

100 ml

Text Solution

Verified by Experts

The correct Answer is:
A
Equivalent mass of metal carbonate `=20+30=50.`
Now, 2.5 gm of metal carbonate `=(2.5)/(50)="0.05 geq"`
Number of equivalents of `H_(2)SO_(4)` that would have reacted = 0.05
Number of equivalents of `H_(2)SO_(4)` taken `=(100xx1)/(1000)="0.1 geq"`
Number of equivalent of `H_(2)SO_(4)` which remained unreacted `= 0.1- 0.05 = "0.05 geq ".`
So,`" "V=(0.05xx1000)/(1.0)=50ml`
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