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Question

# (a) 2 g of a metal carbonate were dissolved in 50 mL of N HCl. 100 mL of 0.1 N NaOH were required to neutralise the resultant solution. Calculate the equivalent mass of the metal carbonate.(b) How much water should be added to 75 mL of 3 N HCl to make it a normal solution?

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Solution

## A) we have,MCO3+2HCl→MCl2+CO2+H2Owe have, 50 ml of 1 N HClno . of moles of HCl = 1∗501000 = 0.05 mole100 ml of 0.1 N NaOH , have no of moles = 0.01molemoles of HCl left after neutralisation with NaOH = 0.05 - 0.01 = 0.04 molethe remaining moles of HCl is reacting with metal carbonate 2 mole of HCl is reacting with 1 mole of metal carbonate so, 0.04 mole of HCl is reacting with metal carbonate = 0.042 = 0.02 molesmoles of metal carbonate = 0.02 = givenweightmolecularweightwe get, molecular weight = 100gB) we have, N1V1=N2V2where, V2 is final volumeso we get, 3.751000=V2∗0.1 V2 = 225 mlamount of water added = 225 - 75 = 145 ml

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