Question

Calculate the weight of anhydrous sodium carbonate required for the preparation of 250ml solution with 0.5N strength​

Answer 1

Answer:

4gram

Explanation:

Here the word anhydrous is given which means no water molecule in this salt

Answer 2

Explanation:

Semi - normal mean

Normality =0.5 N

Now normality =

Volume of solution

no. of moles× n factor/volume of solutions

n− factor of sodium Carbonate =3

{Na2CO3- 2Na+Co3}

Volume of solution =500 ml

=0.5 L

Now,

Normality =0.5 N

⇒

no. of moles×3/0.5 =0.5

⇒ no. of moles = 0.5×0.5/3

= 0.25/3

weight = molar mass × no. of moles

=(23×2+12+3×16)× 0.25/3

=(46+12+48)×0.25/3

=106× 0.25/3 = 53/6

=8.833 g