A sample of urine containing `0.3 g` of urea was treated with an excess of `0.2 M` nitrous acid, according to the equation.
`NH_(2)CONH_(2) + 2HNO_(2) rarr CO_(2) + 2N_(2) + 2H_(2)O`
The gass produced passed through aqueous `KOH` solution and the final valume is measured.
(Given, `Mw_("urea") = 60 g mol^(-1)`, molar volume of gas at standard condition, i.e., at room temperature `25^(@)C` and 1 atm pressure. `RTP` (room temperature pressure) also is `24.4 L` or `24400 mL mol^(-1))`
What is the volume of `HNO_(2)` consumed by urea?

Options

A. 89.6c c
B. 179.2cc
C. 44.8cc
D. 22.4cc

Answer

A

Explanation

`underset(1 "mole")(NH_(2)CONH_(2))+ 2HNO_(2) rarr CO_(2) + underset(2"mole"(or)vol)(2N_(2)) + 3H_(2)O`
Aqueous `NaOH` solution absorbs `CO_(2)`
`60g .......2 xx 22.4`. Its at S.T.P
`0.120g ...........?`