Question

calculate the normality of 0.2(M) aqueous solution of a dibasic acid​

Answer 1

Answer:

0.4N

Explanation:

$\frac{Normality}{molarity}= basicity of acid$

we have 0.2 M solution.

the basicity of acid is 2.

So normality of the solution is 0.4 N

Answer 2

answer : 0.4 N

explanation : you didn't solve this type of question, it means you don't know about molarity and normality.

molarity is the number of mole of solute per litre of solution.

e.g., molarity = mole of solute/volume of solution in L

= mass of solute/volume of solution × molar weight of solute ......(1)

similarly, normality is the gram equivalent of solute per litre of solution.

e.g., normality = gram equivalent of solute/volume of solution.

= mass of solute/{volume of solution × equivalent weight}

we know, equivalent weight = molar weight /n , where n is known as n-factor means change in valency/oxidation for ion/acidity/basicity

so, normality = n × mass of solute/(volume of solution × molar weight of solute) .......(2)

from equation (1) and (2),

normality = n × molarity.

here molarity = 0.2M

and n = 2 [ as dibasic acid ]

so, normality = 2 × 0.2 = 0.4M