nd thus their molalities are equal since the boiling point elevation constant (Kb) and the solvent are the same for both solutions.

The formula for molality (m) is given by:

$$ m = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (kg)}} $$

Since the solutions are dilute and the solvent is water, we can assume 1 $$\mathrm{dm}^3$$ of water (which is 1 kg). Now, we also know that the boiling point elevation per unit molality (Kb) for water is a constant and will not affect our calculations since both solutions have the same change in boiling point.

For the urea solution:

$$ \text{mass of urea} = 3 \mathrm{~g} $$

$$ \text{molar mass of urea} = 60 \mathrm{~g} \mathrm{~mol}^{-1} $$

So, the molality for urea solution is:

$$ m_{\text{urea}} = \frac{3 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1} \times 1 \mathrm{~kg}} = \frac{3}{60} = 0.05 \mathrm{~mol} \mathrm{~kg}^{-1} $$

For the unknown solute A solution:

$$ \text{mass of solute A} = 4.5 \mathrm{~g} $$

Let the molar mass of solute A be $$ M_{\text{A}} $$.

So, the molality for solution A is:

$$ m_{\text{A}} = \frac{4.5 \mathrm{~g}}{M_{\text{A}} \times 1 \mathrm{~kg}} = \frac{4.5}{M_{\text{A}}} \mathrm{~mol} \mathrm{~kg}^{-1} $$

Since the boiling point elevations of both solutions are the same, their molalities are equal:

$$ 0.05 = \frac{4.5}{M_{\text{A}}} $$

Solving for $$ M_{\text{A}} $$:

$$ M_{\text{A}} = \frac{4.5}{0.05} = 90 \mathrm{~g} \mathrm{~mol}^{-1} $$

Therefore, the molar mass of solute A is:

Option D: $$90 \mathrm{~g} \mathrm{~mol}^{-1}$$

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