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

Step-by-step solution

Equivalent weight of metal carbonate =20+30=502.5 ~g of metal carbonate = 2.5 50 =0.05 eq . Number of equivalent of H ₂ SO ₄ would have reacted =0.05 Number of equivalent of H ₂ SO ₄ taken = 100 1 1000 =0.1 Number of equivalent of H ₂ SO ₄ remains unreacted =0.1-0.05=0.05 eq . Number of equivalent of alkali consumed =0.05 eq . milli eq. = Normality Volume in mL aligned & 1.0 V=0.05 1000 & V= 0.05 1000 1.0 =50 ~mL aligned