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.

Options

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

Answer

A

Explanation

Equivalent weight of metal carbonate
`=20+30=50`
2.5 g of metal carbonate `=(2.5)/(50)=0.05eq.`
Number of equivalent of `H_(2)SO_(4)` taken.
`=(100xx1)/(1000)=0.1`
Number of equivalent of `H_(2)SO_(4)` remains unreacted `=0.1-0.05 = 0.05 eq.`
`therefore` Number of equivalent of alkali consumed = 0.05 eq.
`"milli eq. "="Noramlity "xx"Volume in mL"`
`therefore 1.0xxV=0.05xx1000`
`V=(0.05xx1000)/(1.0)="50 ml"`