Question

In a clinical laboratory, a sample of urine containing $0.120$ g of urea $N{H}_{2}CON{H}_2$ (mol. wt. = $60$) was treated with an excess of nitrous acid. The urea reacted according to the following equation:

$N{H}_{2}CON{H}_2+2HN{O}_2\to C{O}_2+2N_2+3H_{2}O$

The gases formed were passed through aqueous sodium hydroxide and the final volume was measured at STP. What was the final volume?

• A. $89.6$ cc
• B. $179.2$ cc
• C. $44.8$ cc
• D. $22.4$ cc

Answer 1

Answer: (A)

$89.6$ cc

Moles of urea used $=\frac{0.120}{60}=0.002$ mol

As per the reaction moles of $C{O}_2$ and $N_2$ gas formed are 0.002 and 0.004 respectively.

The evolved gases were passed through $NaOH$ where $C{O}_2$ got dissolved.

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

Volume of nitrogen gas $=22.4\times 0.004$ L$=0.0896$ L$=89.6$ cc

Hence, option $A$ is correct.