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

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Correct answer: A. 50

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 reacted =0.05Number of equivalent of H_2SO_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 in mL aligned & 1.0 × V=0.05 × 1000 & V=(0.05 × 1000)/(1.0)=50 ~mL aligned

2.5 ~g of the carbonate of a metal was treated with 100 ~mL of 1 ~N...

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