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 at `RTP`?

Options

A. `122 mL`
B. `244 mL`
C. `366 mL`
D. `488 mL`

Answer

B

Explanation

`NH_(2)CONH_(2) + 2HNO_(2) rarr CO_(2) uarr + 2N_(2) uarr + 3H_(2)O(l)`
mmoles `= (0.3 xx 10^(3))/(60)`
`= 5`
Initial 5 10 - - -
Final - - 5 10 15
`V_(H_(2)O)` at room temperature = 0 (since it is liquid).
`V_(CO_(2))` is absorbed by `KOH` or `NaOH`.
`:.` millimoles of gases (i.e., `N_(2)`) = 10
`V` at `RTP = 24400 xx 10 xx 10^(-3) mL = 244 mL`