Step 1: Understanding the Concept:
Two solutions are isotonic when they have the same osmotic pressure (\(\pi_{1} = \pi_{2}\)) at the same temperature.
If both substances are non-electrolytes (or have the same van't Hoff factor), isotonicity implies that their molar concentrations must be equal (\(C_{1} = C_{2}\)).
Molarity is defined as the number of moles of solute per liter of solution.
In percentage solutions (usually % w/v in these problems), we compare the number of moles in a given volume of solution.
Step 2: Key Formula or Approach:
For isotonicity: \(C_{1} = C_{2}\).
Molar concentration \(C = \frac{n}{V} = \frac{w}{M \cdot V}\).
Since we can compare them at the same volume (e.g., 100 mL or 1 L):
\[ \frac{w_{1}}{M_{1}} = \frac{w_{2}}{M_{2}} \]
Where:
\(w_{1}, w_{2}\) are the mass percentages (or mass in the same volume).
\(M_{1}, M_{2}\) are the molecular weights.
Step 3: Detailed Explanation:
Let's assign the variables:
1. Cane Sugar (Solution 1):
Mass percentage (\(w_{1}\)) = 5% (meaning 5g of sugar in 100 mL).
Molecular weight (\(M_{1}\)) = 342.
2. Substance X (Solution 2):
Mass percentage (\(w_{2}\)) = 1% (meaning 1g of substance X in 100 mL).
Molecular weight (\(M_{2}\)) = Unknown.
Substitute these values into the equality:
\[ \frac{5}{342} = \frac{1}{M_{2}} \]
To find \(M_{2}\), perform cross-multiplication:
\[ 5 \cdot M_{2} = 342 \cdot 1 \]
\[ M_{2} = \frac{342}{5} \]
\[ M_{2} = 68.4 \].
Therefore, for the 1% solution of substance X to have the same number of particles (moles) as the 5% sugar solution, its molecules must be much lighter, specifically 68.4 g/mol.
Step 4: Final Answer:
The molecular weight of substance X is 68.4.