Question

0.4 moles of Na2CO3 is present in500 ml of soln. Normality will be

Answer 1

Moles of Na2CO3=0.4

Volume of solution=500ml=0.5litre

N Factor of Na2CO3 =2

Normality (N)= Molarity (M)×n Factor

M=mole of Na2CO3/vol

M=0.4/0.5=0.8

N=M×n Factor

N=0.8×2

Normality=1.6N

I hope this helps ( ╹▽╹ )

Answer 2

Explanation:

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