2.5 g of the carbonate of a metal was 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.

Options

A. 50 ml
B. 25 ml
C. 75 ml
D. 100 ml

Answer

A

Explanation

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`