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. (a)`50`
B. (b)`25`
C. (c )`75`
D. (d)`100`

Answer

d

Explanation

Equivalent weight of metal carbonate
`=20+30=50`
`2.5 g` of metal carbonate `=2.5/50=0.05 eq.`
Number of equivalent of `H_(2)SO_(4)` would have reached `=0.05`
Number of equivalent of `H_(2)SO_(4)` taken
`=(100xx1)/1000=0.1`
`:.` Number of equivalent of `H_(2)SO_(4)` remains unreached
`=0.1-0.05=0.05 eq`.
`:.` Number of equivalent of alkali consumed `=0.05 eq.`
Milli eq. =Normality `xx` Volume in mL.
`:. 1.0xxV=0.05xx1000`
`V=(0.05xx1000)/1.0=50 mL`