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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`?A. `122 mL`B. `244 mL`C. `366 mL`D. `488 mL` |
Answer» Correct Answer - B `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` |
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