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A sample of urine containing 0.3 g of ur...

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?

A

`12.5 mL`

B

`25 mL`

C

`50 mL`

D

`75 mL`

Text Solution

Verified by Experts

The correct Answer is:
C
millimoles of `HNO_(2) = 10 = 0.2 M xx V_(HNO_(2)) (mL)`
`V_(NHO_(2)) = 50 mL`
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