A solution containing $10 \text{ g/dm}^3$ of urea (molecular mass $= 60 \text{ g mol}^{-1}$) is isotonic with a $5 \%$ solution of a non-volatile solute. The molecular mass of this non-volatile solute is:

Two solutions are said to be isotonic if they have the same osmotic pressure ($\pi$) at a given temperature . The osmotic pressure is given by the formula $\pi = CRT$, where $C$ is the molar concentration .

For isotonic solutions: $\pi_1 = \pi_2 \implies C_1RT = C_2RT \implies C_1 = C_2$

1. Calculate Concentration of Urea ($C_1$): Concentration given = $10 \text{ g/dm}^3 = 10 \text{ g/L}$. Molar mass of urea = $60 \text{ g mol}^{-1}$.

• $C_1 = \frac{10}{60} \text{ mol L}^{-1}$.

2. Calculate Concentration of Unknown Solute ($C_2$): Concentration given = $5 \%$. In the context of osmotic pressure problems where density is not provided, this is interpreted as mass/volume percentage ($w/v$), meaning $5 \text{ g}$ solute in $100 \text{ mL}$ solution. Concentration in g/L = $5 \text{ g} \times \frac{1000 \text{ mL}}{100 \text{ mL}} = 50 \text{ g/L}$. Let the molecular mass of the unknown solute be $M_2$. $C_2 = \frac{50}{M_2} \text{ mol L}^{-1}$.

3. Equate and Solve: $\frac{10}{60} = \frac{50}{M_2}$ $M_2 = \frac{50 \times 60}{10} = 300 \text{ g mol}^{-1}$.