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

1.	5 gm of Na2co3 is dissolved in 200 ml of the solution.calculate normality of the solution.
Answer» 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.

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

Answer»

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.