JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Some Basic Concepts of Chemistry

Given:

• Mole fraction of urea, $X_{\text{urea}}=0.05$
• Mass of water = 900 g → moles of water = $\frac{900}{18}=50$ mol

Let moles of urea = $n$

$X_{\text{urea}}=\frac{n}{n+50}=0.05$

Solving: $n=0.05(n+50)$
$n=0.05n+2.5$
$0.95n=2.5$
$n=\frac{2.5}{0.95}=2.6316$ mol

Mass of urea = $2.6316\times 60=157.89$ g

Total mass of solution = $900+157.89=1057.89$ g

Volume of solution = $\frac{1057.89}{1.2}=881.58$ mL = 0.88158 L

Molarity = $\frac{2.6316}{0.88158}=2.985$ M ≈ 2.98 M

(The source lists two possible answers: 2.80 or 3.05. My calculation gives 2.98, which falls between these values but doesn't match either exactly. This may indicate a calculation variation or rounding difference in the source.)