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The ammonia evolved from the treatment of 0.30g of an organic compound for the estimation of nitrogen was passed in 100ml of 0.1 M H_(2)SO_(4). The excess acid required 20ml of 0.5M NaOH solution for complete neutralisation . The organic compound is: |
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Answer» Thiourea `underset("acid")(N_(1)V_(1))=underset("base")(N_(2)V_(2))` `0.2xxV_(1)=0.5xx20` `V_(1)=(0.5xx20)/(0.2)=50mL` Volume of `H_(2)SO_(4)` used `=100-50=50mL` `%` of `N` in the COMPOUND `=(1.4xxVxxN)/(W)` `=(1.4xx50xx0.2)/(0.30)=46.6%` The PERCENTAGE of N is same as in urea `(NH_(2)CONH_(2))` Percentage of `N=("MASS of "N)/("Mass of urea")xx100` `=(14xx2)/(60)xx100=46.6%` |
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