1. Use Raoult’s law for a non-volatile solute

For urea dissolved in water, urea is non-volatile, so:

$P = X_{\text{water}} P^0$

Hence, relative lowering of vapour pressure is:

$\frac{P^0 - P}{P^0} = 1 - X_{\text{water}} = X_{\text{solute}}$

Given vapour pressure is reduced by $25\%$:

$\frac{P^0 - P}{P^0} = 0.25$

So,

$X_{\text{urea}} = 0.25$

2. Calculate moles of water

Mass of water $= 1000\,\text{g}$

Molar mass of water:

$18\,\text{g mol}^{-1}$

Therefore,

$n_{\text{water}} = \frac{1000}{18} = 55.56\,\text{mol}$

3. Set up mole fraction equation

Let moles of urea be $n$.

Then,

$X_{\text{urea}} = \frac{n}{n + 55.56} = 0.25$

Solve:

$n = 0.25(n + 55.56)$

$n = 0.25n + 13.89$

$0.75n = 13.89$

$n = 18.52\,\text{mol}$

4. Calculate molar mass of urea

Urea: $\mathrm{NH_2CONH_2}$

This contains:

• $2$ N atoms
• $1$ C atom
• $1$ O atom
• $4$ H atoms

So molar mass:

$M = 2(14) + 12 + 16 + 4(1) = 28 + 12 + 16 + 4 = 60\,\text{g mol}^{-1}$

5. Calculate mass of urea required

$m = nM = 18.52 \times 60 = 1111.2\,\text{g}$

Nearest integer:

$\boxed{1111}$

6. Compare with stored correct answer

Derived answer = $1111$

Stored correct answer = $1111$

They match.