Isotonic Solutions Principle

Two solutions are isotonic if they have the same osmotic pressure. For solutions containing non-electrolytes, the osmotic pressure ($\pi$) is given by $\pi = CRT$, where $C$ is the molar concentration, $R$ is the ideal gas constant, and $T$ is the absolute temperature. If two solutions are isotonic, their molar concentrations ($C$) must be equal under the same temperature conditions, assuming they contain non-electrolytes (van't Hoff factor $i=1$ for both).

In this case, the Urea solution is isotonic with the 5% solution of solute A. Both are treated as non-electrolytes.

Therefore, the molarity of the Urea solution must equal the molarity of the solute A solution: $C_{\text{Urea}} = C_{\text{A}}$

Urea Solution Molarity Calculation

Given:

• Concentration of Urea = 8.3 g/dm$^3$ (which is equivalent to 8.3 g/L)
• Molar mass of Urea ($M_{\text{Urea}}$) = 60 g mol$^{-1}$

The molarity ($C_{\text{Urea}}$) is calculated as:

Solute A Solution Molarity Calculation

Given:

• Concentration of Solute A = 5% solution. Assuming this is a 5% w/v (mass/volume) solution, it means 5 g of solute A per 100 mL of solution.
• This concentration can be expressed as 50 g L$^{-1}$ (since 5 g / 100 mL = 50 g / 1000 mL = 50 g / L).
• Let the molar mass of Solute A be $M_{\text{A}}$ (in g mol$^{-1}$).

The molarity ($C_{\text{A}}$) is calculated as:

Equating Molarities and Finding Molar Mass

Using the isotonic condition ($C_{\text{Urea}} = C_{\text{A}}$):

Now, solve for $M_{\text{A}}$:

The approximate molar mass of solute A is 361.5 g mol$^{-1}$.