Question

A $ 4\% $ solution of a non-volatile solute is isotonic with $ 0.702\% $ urea solution. Calculate the molar mass of the non-volatile solute. (Molar mass of urea is $ 60{\text{g/mol}} $ )

Answer 1

Hint
In this question, the concept of the isotonic solution will be used, that is the Isotonic solution means the solution has the same osmotic pressure. Use the concept of the osmotic pressure to obtain the result.

Complete step by step answer
Osmotic pressure is a minimum pressure which is needed to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. It is also defined as the measure of the tendency of a solution to take in pure solvent by the osmosis process.
Here in this case, $ 0.702\% $ urea solution means $ 0.702 $ g urea is dissolved in $ 100 $ ml of solvent to make it a solution. Also $ 4\% $ solution of nonvolatile solute means $ 4 $ g of nonvolatile solute is dissolved in $ 100 $ ml solvent.
So as we have the weight of urea, we first calculate the number of moles of urea which is-
 $ \Rightarrow {\text{Number of moles of urea}} = \dfrac{{0.702}}{{60}} = {\text{ }}0.0117 $
Again we know that Osmotic Pressure is mathematically written as-
 $ \Rightarrow P = \pi NRT $
Where $ N $ is the number of moles, $ R $ is the gas constant, $ T $ is the temperature of the solution.
In this case the osmotic pressure due to the nonvolatile solute and urea will be the same.
Let us consider, the osmotic pressure due to nonvolatile solute is $ {P_1} $ , the number of moles of nonvolatile solute is $ {N_1} $ , and the osmotic pressure due to urea is $ {P_2} $ , the number of moles of urea is $ {N_2} $
For the same pressure we can write,
 $ \Rightarrow \pi {N_1}RT = \pi {N_2}RT $
 $ \Rightarrow {N_1} = {N_2} $
 $ \Rightarrow {N_2} = {\text{ }}0.0117 $
Let the Molar mass of the solute be $ M $
 $ \Rightarrow {\text{Number of moles}} = {\text{ }}\dfrac{{\left( {weight} \right)}}{M} $
Now, we substitute the values in the above equation as,
 $ \Rightarrow 0.0117 = \dfrac{4}{M} $
After simplification we get,
 $ \Rightarrow M{\text{ }} = {\text{ }}341.88{\text{ g/mol}} $
Hence, the Molar Mass of the non-volatile solute is $ 341.88{\text{ g/mol}} $.

Note
Osmosis is the rapid movement of solvent molecules through a specific permeable membrane into a region of higher solute concentration, in the direction that tends to equalize the solute concentrations on both the sides.