Question

A 5% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol\(^{-1}\)) in the same solvent. If densities of both solutions are \(1 \, \text{g cm}^{-3}\), the molar mass of the substance is _ _ _ _ _ .

Hint:

For isotonic solutions with same density, directly equate molarity after converting % to molarity.

Answer 1

Correct Answer: 200

Concept: Isotonic solutions have same osmotic pressure: \[ \pi = CRT \Rightarrow C_1 = C_2 \]

Step 1: Density = \(1 \, \text{g cm}^{-3}\) \(\Rightarrow\) \(100 \, \text{mL}\) solution = \(100 \, \text{g}\) Urea solution: \[ 1.5\% \Rightarrow 1.5 \, \text{g urea in 100 mL} \] \[ \text{Moles of urea} = \frac{1.5}{60} = 0.025 \] \[ \text{Molarity} = \frac{0.025}{0.1} = 0.25 \, M \]

Step 2: Unknown substance solution: \[ 5\% \Rightarrow 5 \, \text{g substance in 100 mL} \] Let molar mass = \(M \, \text{g mol}^{-1}\) \[ \text{Moles} = \frac{5}{M} \] \[ \text{Molarity} = \frac{5/M}{0.1} = \frac{50}{M} \]

Step 3: For isotonic solutions: \[ \frac{50}{M} = 0.25 \] \[ M = \frac{50}{0.25} = 200 \, \text{g mol}^{-1} \]