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

Options

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

Answer

C

Explanation

Normality `= n xx` Molarity (`n = 2` for `H_(2)SO_(4), n = 1` for `NaOH`)
Total milliequivalent of `NH_(3) =` (Total Milliequivalent of `H_(2)SO_(4)) -` (Milliequivalent of `H_(2)SO_(4)` neutralized by `NaOH`)
`= (20) - (10) = 10`
Equivalent of `NH_(3) = ("Milliequi valent")/(1000) = (10)/(1000) = 10^(-2)`
`:.` Mass of `N = (n_(N))("Molar mass")`
`= (10^(-2))(14) = 0.14 g`
`% N = (m_(N))/(m_(o .c.)) xx 100% = (0.149)/(0.30 g) xx 100%`
`= 46.6%`
now, we calculate the percentage of `N` in every compound to fix the answer.
Thiourea `(H_(2)NCSNH_(2)) : %N = (28)/(76) xx 100% = 31.84`
benzamide `(C_(6)H_(5)CONH_(2)): % N = (14)/(121) xx 100% = 11.5`
Urea `(H_(2)NCONH_(2)): % N = (28)/(60) xx 100% = 46.6`
Acetamide `(CH_(3)CONH_(2)): % N = (14)/(59) xx 100% = 23.7`
Hence, the given organic compound is urea.