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 of0.1 `MH_2SO_4` be left unused. Then using molarity equation.
`M_1 xx V_1 xxn_1 (H_2SO_4)= M_2 xxV_2 xx n_2(NaOH)`
`V_1 xx0.1 xx 2 = 20 xx 0.5 xx1 = 50 mL`
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_2CONH_2) = 28/60xx100= 46.6%`
(as 28 is the molecular mass of N, and 60 is the molecular mass of urea). This is equal to the percentage of nitrogen obtained above. Thus, the compound present is urea.