The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is

Options

A. acetamide
B. thiourea
C. urea
D. benzamide

Answer

C

Explanation

Let `V_(1)` mL of 0.1 M `H_(2)SO_(4)` be left unused. Then using molarity equation.
`M_(1)xxV_(1)xxn_(1)(H_(2)SO_(4))=M_(2)xxV_(2)xxn_(2)(NaOH)`
`V_(1)xx0.1xx2=20xx0.5xx1=50mL`
Volume of acid used = (100-50)mL = 50 mL of 0.1 M `H_(2)SO_(4)`
Percentage of nitrogen present in the given organic compound = `(1.4xx2xx50xx0.1)/(0.3)=46.6%`
Nitrogen present in urea `(NH_(2)CONH_(2))=(28)/(60)xx100=46.6%`
(as 28 is the molecular mass of `N_(2)` and 60 is the molecular mass of urea). This is equal to the percentage of nitrogen obtained above. Thus, the compound present in urea.