Question

Find the amount of n a 2 c o 3 required to prepare 250 ml of n by 20 n a 2 c o 3 solution

Answer 1

normality=1/20

here, n-factor =2

we know, N=nM

where N= normality and M= molarity

M= number of moles/volume of solution

N=n×number of moles/volume of solution

number of moles=mass/ mol. wt

let mass be m, mol wt of Na2CO3 = 122

hence N=(2×m×1000)/(122×250)

m=122×250/20×1000×2

m=0.7625 g