1.

In a sample of sodium carbonate some sodium sulphate is also mixed. 1.25 g of this sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralises 20 mL of `(N)/(10)` sulphuric acid. Calculate the percentage of sodium carbonate in the sample.

Answer» 25 mL of sample solution neutralises
`20 mL(N)/(10)H_(2)SO_(4)`
250 mL of sample solution will neutralise
`=200 mL (N)/(10)H_(2)SO_(4)`
`200 mL (N)/(10)H_(2)SO_(4)-=200 mL(N)/(10)Na_(2)CO_(3)` solution
Amount of `Na_(2)CO_(3)` present `=(ExxNxxV)/(1000)`
`=(53xx200)/(10xx1000)=1.06`
% of `Na_(2)CO_(3)` in the sample `=(1.06)/(1.25)xx100=84.8`


Discussion

No Comment Found

Related InterviewSolutions