Question

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

• A. \(200\, g\, mol^{-1}\)
• B. $300 \,g \,mol^{-1}$
• C. $400 \,g \,mol^{-1}$
• D. $500\, g\, mol^{-1}$

Answer 1

The Correct Option is B

To find the molecular mass of the non-volatile solute, we can use the concept of isotonic solutions. Two solutions are isotonic when they have the same osmotic pressure. The osmotic pressure depends on the concentration (molarity) and the nature of the solute (i.e., number of particles, for non-electrolytes it is just one).

Given:

• Urea solution concentration: 10 \, g\, dm^{-3}
• Molecular mass of urea: 60 \, g \, mol^{-1}
• Non-volatile solute solution: 5\% by weight

First, calculate the molarity of the urea solution:

• Moles of urea in 1 \, dm^3 = \frac{10}{60} \, mol = \frac{1}{6} \, mol
• Molarity, M_{urea} = \frac{1/6}{1} \, mol \, dm^{-3}

For the non-volatile solute (solution is 5% by weight):

• This means 5 \, g of solute is present in 100 \, g of solution.
• Assume the density of the solution is approximately 1 \, g \, cm^{-3}, then 100 \, g of solution is approximately 100 \, cm^3 or 0.1 \, dm^3.
• Moles of non-volatile solute = \frac{5}{M} \, mol, where M is the molar mass of the solute.
• Molarity, M_{solute} = \frac{5/M}{0.1} = \frac{50}{M} \, mol \, dm^{-3}

Since both solutions are isotonic, their molarities must be the same:

• \frac{1}{6} = \frac{50}{M}

Solving for M gives:

• M = 50 \times 6 = 300 \, g \, mol^{-1}

Thus, the molecular mass of the non-volatile solute is 300 \, g \, mol^{-1}. Therefore, the correct answer is:

300 \, g \, mol^{-1}