Calculate the mass of `Na_(2)CO_(3)` required to prepare 250 ml of 0.5 N solution.

Explanation

Normality of required solution = 0.5 N
Volume of required solution = 250 ml
Equivalent wt.of. `Na_(2)CO_(3)=(106)/(2)=53`
Normality (N) = `(W)/(GEW)xx(1000)/(V(mL)`
wt.of solute = N x xGEW xx `(V(ml))/(1000)`
=0.5x x 53x x `(250)/(1000)=(53)/(8)=6.62 g`