3.0 g of impure `Na_2CO_3` is dissolved in water and the solution is made up to `250 mL`. To 50 " mL of " this solution, 50 " mL of " 0.1 N HCL is added and the mixture, after shaking well, required 10 " mL of " 0.16 N NaOH solution for cojmplete neutralisation. Calculate the percentage purity of the sample of `Na_2CO_3`.

Explanation

Excess m" Eq of "HCl=m" Eq of "NaOH
`=0.16xx10=1.6`
m" Eq of "HCl added`=0.1xx50=5.0`
`impliesmEq` of HCl used against `Na_2CO_3=5.0-1.6=3.4`
Also m" Eq of "`Na_2CO_3=5.0-1.6=3.4` (in 50 mL)
`-=17` (in 250 mL)
`implies(g)/((106)/(2))xx1000=17impliesg=0.901`
`%` purity`=(0.901)/(3)xx100=30.03`