2.5 g of the carbonate of metal was treated with 100 ml of 1 N H2SO4. After the completion of the reaction, the solution was boiled off to expel CO2 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 is:

Equivalent weight of metal carbonate=20+30=50
2.5 g of metal carbonate=2.5/50 eq
Number of equivalent of H2SO4 would have reacted=0.05
Number of equivalent of H2SO4 taken=100×1/1000=0.1
Number of equivalent of H2SO4 remains unreacted=0.1−0.05=0.05 eq.
∴Number of equivalent of alkali consumed=0.05 eq.
milli eq.=Normality×Volume in ml
∴1.0×V=0.05×1000
V=0.05×1000/1.0=50 ml