1. Use isotonic condition

For isotonic solutions at the same temperature: $\pi = C R T$ So, if both are isotonic in the same solvent and both solutes are non-electrolytes, $C_1 = C_2$ where $C$ is molarity.

2. Interpret percentage solutions using density

Given density of both solutions is $1.0\ \text{g cm}^{-3}$, so: $100\ \text{cm}^3 = 100\ \text{mL}$ has mass $100\ \text{g}$.

Thus:

• A $5.25\%$ solution contains $5.25\ \text{g}$ solute in $100\ \text{mL}$ solution.
• A $1.5\%$ urea solution contains $1.5\ \text{g}$ urea in $100\ \text{mL}$ solution.

3. Find molarity of urea solution

Moles of urea in $100\ \text{mL}$: $n=\frac{1.5}{60}=0.025\ \text{mol}$

Volume of solution: $100\ \text{mL}=0.1\ \text{L}$

So molarity of urea solution is: $M=\frac{0.025}{0.1}=0.25\ \text{mol L}^{-1}$

4. Apply isotonic condition to unknown solution

The unknown solution is also $0.25\ \text{M}$.

In $100\ \text{mL}$ ($=0.1\ \text{L}$), moles of unknown solute are: $n=0.25\times 0.1=0.025\ \text{mol}$

Given mass of unknown solute in $100\ \text{mL}$ is $5.25\ \text{g}$.

Therefore molar mass is: $\text{Molar mass} = \frac{5.25}{0.025}=210\ \text{g mol}^{-1}$

5. Check options

$210\ \text{g mol}^{-1}$ matches Option D.

Final Answer:

$\boxed{210\ \text{g mol}^{-1}}$