Question

Q. What weight of
Naco3 would be required
to prepare 500 ml
o.1m solution?​

Answer 1

Explanation:

Semi - normal mean

Normality =0.5 N

Now normality =

Volume of solution no. of moles× n factor

n− factor of sodium Carbonate =3

{Na 2CO 3 ⟶2Na + +CO 32− }

Volume of solution =500 ml

=0.5 L

Now,

Normality =0.5 N

⇒ 0.5

no. of moles×3

=0.5

⇒ no. of moles = 30.5×0.5

= 30.25

weight = molar mass × no. of moles

=(23×2+12+3×16) × 30.25

=(46+12+48)× 30.25,

=106× 30.25

= 653

=8.833 g

I hope it's help you..❤️❤️✌️