Question

To prepare 250 ml of a n/10 na2co3 solution solution

Answer 1

First we have to calculate,

molecular weight of Na2CO3 = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106 g

Equivalent weight of Na2CO3 = 106 / 2 = 53 g

Now, we need to calculate the quantity of Na2CO3 to prepare the given solution. Let us assume it to be x.

So, the formula is as follows:

Normality = (weight of salt / equivalent weight) × (1000 / volume in ml )

Therefore, 0.1 = (x / 53) × (1000 / 250)

x = (0.1 × 53 × 250) / 1000

x= 1.325 g

So, we have to add 1.325 g of Na2CO3 in the distilled water, stir it well and make up the final volume to 250 ml.