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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. |
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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` |
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