Question

$5.0g$ sample of urea when heated with $NaOBr$ and $NaOH$ gives $1120ml$ of nitrogen at STP. The percentage purity of the sample is:

• A. $50$%
• B. $40$%
• C. $60$%
• D. $30$%

Answer 1

The correct option is C $60$%

${NH}_{2}CO{NH}_2+3NaOBr+2NaOH⟶N_2+3NaBr+{Na}_2{CO}_3+3H_{2}O$

$\because$ $22400ml$ of $N_2$ is obtained from $60g$ of urea at S.T.P.

$\therefore$ $1120ml$ of $N_2$ will be obtained from

$\Rightarrow \frac{60\times 1120}{22400}g$ of urea $=3g$ or urea

Now,
$\because$ $5g$ sample contains $3.0g$ of urea

$\therefore$ $100g$ sample will contain $\frac{3.0\times 100}{5.0}=60g$ of urea

Hence, the purity of urea sample is $60$%