The density of a solution prepared by dissolving $$120$$ g of urea (mol. mass $$=60u$$) in $$1000$$ g of water is $$1.15$$ g/mL. The molarity of this solution is :

Molarity $$M$$ is defined as “number of moles of solute present in one litre of the solution”.
Therefore we need:

1. number of moles of urea (solute)
2. total volume of the resulting solution (in litres)

Step 1: Moles of urea
Given mass of urea $$=120\,\text{g}$$, molar mass $$=60\,\text{g mol}^{-1}$$.
Number of moles $$n=\dfrac{\text{mass}}{\text{molar mass}}=\dfrac{120}{60}=2\,$$mol.

Step 2: Mass of the whole solution
Mass of water $$=1000\,\text{g}$$.
Mass of urea $$=120\,\text{g}$$.
Total mass of solution $$=1000+120=1120\,\text{g}$$.

Step 3: Volume of the solution
Density $$\rho=1.15\,\text{g mL}^{-1}$$.
$$\text{Volume}=\dfrac{\text{mass}}{\rho}= \dfrac{1120\,\text{g}}{1.15\,\text{g mL}^{-1}} \approx 973.9\,\text{mL}$$.
Convert to litres: $$973.9\,\text{mL}=0.9739\,\text{L}\approx0.974\,\text{L}$$.

Step 4: Molarity of the solution
$$M = \dfrac{n}{V} = \dfrac{2.00\,\text{mol}}{0.974\,\text{L}} \approx 2.055\,\text{mol L}^{-1}$$.

Rounding to three significant figures gives $$2.05\,\text{M}$$.

Hence the molarity of the solution is $$2.05\,\text{M}$$.
Option D which is: $$2.05$$ M