What weight of CaCO3 shall be neutralised by 100 ml of 0.1 M(molarity) HCl?

For neutralisation , No. Of equivalent of CaCO3 = no. Of equivalent of HClNo. Of equivalent of HCL = normality × volume =molarity × valency factor × volume = 0.1 × 1 × 0.1 = 0.01 Now equivalent of calcium carbonate should be equal to equivalent of HCL0.01= wt /equivalent wt ( equivalent wt = molecular wt / valency factor ) = 100 /2 = 50 0.01 = wt / 50Wt = 0.5 gm

No of equvelence of HCl = N×M×V = 0.01No. of equvelence of CaCO3= (wt taken) /equvelence wtequvelence wt of CaCO3= molecular wt/