The ammonia evolved from the treatment of 0.03 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The axcess of the acid required 20 mL of 0,5 M solution hydroxide solution for complete neutralisation. the organic compound is :

Options

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

Answer

C

Explanation

Le the vol. of acid left unused
= v mL of 0.1 `M H_(2)SO_(4)`
Applying molarity equation
`n_(a)M_(a)V_(a) = n_(b)M_(b)V_(b)`, we have,
`2 xx 0.1 xx v = 1 xx 0.5 xx 20` or v = 50 mL
`:.` Vol. of acid used = 100 - 50
= 50 mL of `0.1 M H_(2)SO_(4)`
`%N = (1.4 xx n_(1)M_(a)V_(a))/("wt. of substance taken") = (1.4 xx 2 xx 0.1 xx 50)/(0.3)`
`= 46.6`
% N in urea `(NH_(2)CONH_(2)) = (28//60) xx 100 = 46.6%`, in acetamide `(CH_(3)CONH_(2)) = (14//59) xx 100 = 23.72%`, in benzamide `(C_(6)H_(5)CONH_(2)) = (14//21) xx 100 = 11.57%` and in thiourea `(NH_(2)CSNH_(2)) = (28//76) xx 100 = 36.84%`
Thus, option (c) is correct.