1.15g of an organic compound was analysed by Kjeldahl's method and ammonia produced was collected in 30 mL of normal Hcl solution. the excess of acid consumed 18.4 mL of normal sodium hydroxide solution for back titration. calculate the percentage of nitrogenin the substance
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the required amount of acid is (30-18.4)mL = 11.6 mL 1N
now 1mL 1N HCl≡1mL 1N NH3
∴11.6 mL 1N HCl≡11.6 mL 1N NH3
now 1000mL 1N NH3≡14.0g Nitrogen
11.6mL 1N NH3≡(14*11.6)/1000g Nitrogen=0.1624g nitrogen
∴ %N=(0.1624/1.15)*100=14.12
now 1mL 1N HCl≡1mL 1N NH3
∴11.6 mL 1N HCl≡11.6 mL 1N NH3
now 1000mL 1N NH3≡14.0g Nitrogen
11.6mL 1N NH3≡(14*11.6)/1000g Nitrogen=0.1624g nitrogen
∴ %N=(0.1624/1.15)*100=14.12
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