Question:
A 5% solution of cane sugar (molar mass 342) is isotonic with 1% solution of non-electrolyte substance X, the molar mass of substance X is
A 5% solution of cane sugar (molar mass 342) is isotonic with 1% solution of non-electrolyte substance X, the molar mass of substance X is
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For isotonic solutions, equating the osmotic pressures allows you to find the molar mass of an unknown solute.
Updated On: Feb 18, 2026
- 34.2 g mol$^{-1}$
- 171.2 g mol$^{-1}$
- 136.8 g mol$^{-1}$
- 68.4 g mol$^{-1}$
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The Correct Option is D
Solution and Explanation
Step 1: Using the concept of isotonic solutions.
Two solutions are isotonic when they have the same osmotic pressure. The osmotic pressure depends on the molar concentration of the solute, and in this case, the solutions of cane sugar and substance X must have the same osmotic pressure.
Step 2: Osmotic pressure calculation.
Osmotic pressure is given by: \[ \Pi = \frac{n}{V}RT \] For isotonic solutions, the ratio of the molar mass of substance X to that of cane sugar can be used to find the molar mass of X. By equating the osmotic pressures, we get the molar mass of substance X as 68.4 g mol$^{-1}$.
Step 3: Conclusion.
The correct answer is (D) 68.4 g mol$^{-1$}, as this is the molar mass of the non-electrolyte substance X.
Two solutions are isotonic when they have the same osmotic pressure. The osmotic pressure depends on the molar concentration of the solute, and in this case, the solutions of cane sugar and substance X must have the same osmotic pressure.
Step 2: Osmotic pressure calculation.
Osmotic pressure is given by: \[ \Pi = \frac{n}{V}RT \] For isotonic solutions, the ratio of the molar mass of substance X to that of cane sugar can be used to find the molar mass of X. By equating the osmotic pressures, we get the molar mass of substance X as 68.4 g mol$^{-1}$.
Step 3: Conclusion.
The correct answer is (D) 68.4 g mol$^{-1$}, as this is the molar mass of the non-electrolyte substance X.
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