Question

Find the normality of the solution containing 5.300 g/L of Na2CO3 ?

Answer 1

Normality = 0.1 N

Step-by-step explanation:

Data given:

weight of Na₂CO₃ = 5.3g/L

Normality: ?

Normality:

It is the expression of concentration for a solution.

Normality is the gram equivalent of a compound in the per liter of solution.

Equivalent:

one equivalent can react with hydrogen ion or replace hydrogen ion.

Formula for Normality:

Normality = gram equivalent of solute / volume of solution in L ....(1)

we know

gram equivalent of solute = weight/ equivalent weight ....... (2)

So by combining equation 1 and 2 the formula will be

N = weight / equivalent weight / volume of solution in L ........ (3)

Now first we have to find equivalent weight.

As in Na₂CO₃ there are 2 Na that can replace hydrogen ion in a solution,

So, we have 2 equivalent

The Molar mass of the Na2CO3 = (23x2 + 12 + 16x3)

Na2CO3 = 106 g/mol

Now to find equivalent weight:

equivalent weight: molar mass / no. of equivalent (n)

equivalent weight: 106 g/mol / 2

equivalent weight: 53 g/mol

Put the values in Equation 2:

gram equivalent = weight/ eq. weight

gram equivalent = 5.3 / 53

gram equivalent = 0.1

Now put this value of gram equivalent in equation 1

Normality = gram equivalent of solute / volume of solution in L

Normality = 0.1 /1L

Normality = 0.1 N

So the Normality of the solution is 0.1N