Question

5 gm of Na2co3 is dissolved in 200 ml of the solution.calculate normality of the solution.​

Answer 1

Answer:

0.5 N

Explanation:

molecular weight of Na2CO3 is = 106

equivalent weight of Na2CO3 is = 106/2 = 53

now, 5 g of Na2CO3 means , (5/53)=0.094 equivalence of Na2CO3

hence,

200mL of solution contains 0.094 equivalence of salt.

then, followed by unitary method we get,

1000 mL of solution contains

(0.094*1000)/200 equivalence of Na2CO3

or, 0.47≈0.5 equivalence of Na2CO3

now notice carefully, it's the Normality of solution.