2.5 g of a carbonate of a metal was treated with 100ml of 1N sulphuric acid after the completion of the rxn the solution was boiled off to expel carbon-dioxide and was then titrated against 1N NaOH solution .the volume of alkali that would be consumed if the equivalent weight of the metal is 20

Dear student ,
   as it is given that 1N sulphuric acid will react with metal carbonate and   sodium hydroxide both one by one . let V ml of  sulphuric acid reacts with  metal carbonate then (100 -V ) ml with sodium hydroxide .
therefore ;  no. of gm equi. of  metal carbonate = no. of gm equi. of V ml of sulphuric acid
                          wt.(in gm) / eq.wt of  metal carbonate  =  N $\times$ V (in L)
                            2.5 / (20+30)   = 1$\times$V $\times$ 10^-3
                                             V     =  50 ml ,therefore vol. of  acid left = 50 ml
now for the reaction between acid and base : N₁ V₁ = N₂ V₂  $\Rightarrow$ 1 $\times$50 = 1 $\times$ V₂  $\Rightarrow$ V₂ = 50 ml
                                                                        (acid)      (base)