Question

In a clinical laboratory, a sample of urine containing 0.120g of urea
NH,CONH, (M.Wt.60). was treated with excess of nitrous acid. The urea reacted
according to the following equation NH CONH, + 2HNO, CO2+2N2 +3H20. The
gas formed was passed through aqueous sodium hydroxide and final volume is
measured at STP, what was this volume.
​

Answer 1

Answer:

Laboratory themometer

Answer 2

Answer:

89.6cc or 0.896L

Explanation:

Moles of urea used =600.120​=0.002 mol

As per the reaction moles of CO2​ and N2​ gas formed are 0.002 and 0.004 respectively.

The evolved gases were passed through NaOH where CO2​ got dissolved.

So, from the reaction, 0.004 moles of nitrogen gas were left.

Volume of nitrogen gas =22.4×0.004 L=0.0896 L=89.6 cc