Question

A 1% solution of solute ‘X’ is isotonic with a 6% solution of sucrose (molar mass = \( 342 \) g mol\(^{-1}\)). The molar mass of solute ‘X’ is:

• A. \( 34.2 \) g mol\(^{-1}\)
• B. \( 57 \) g mol\(^{-1}\)
• C. \( 114 \) g mol\(^{-1}\)
• D. \( 3.42 \) g mol\(^{-1}\)

Hint:

For isotonic solutions, use the formula \( \frac{w_1}{M_1} = \frac{w_2}{M_2} \). If one solution is a percentage solution, express it as grams per 100 mL.

Answer 1

The Correct Option is B

Step 1: Understanding Isotonic Solutions Two solutions are isotonic if they have the same osmotic pressure, which is given by the formula: \[ \Pi = \frac{w}{M} \times \frac{1000}{V} \] where:
- \( \Pi \) = osmotic pressure,
- \( w \) = mass of solute in grams,
- \( M \) = molar mass of solute,
- \( V \) = volume of solution in mL.
For isotonic solutions, their osmotic pressures are equal: \[ \frac{w_1}{M_1} = \frac{w_2}{M_2} \] Step 2: Applying Given Data
- Sucrose solution: \( 6% \) solution means \( 6 \) g of sucrose in \( 100 \) mL solution.
- Solute ‘X’ solution: \( 1% \) solution means \( 1 \) g of solute ‘X’ in \( 100 \) mL solution.
- Molar mass of sucrose: \( M_1 = 342 \) g mol\(^{-1}\).
- Molar mass of solute ‘X’: \( M_2 \) (to be determined).
Using the isotonic equation: \[ \frac{6}{342} = \frac{1}{M_2} \] Step 3: Solving for \( M_2 \) Rearranging: \[ M_2 = \frac{1 \times 342}{6} \] \[ M_2 = 57 \text{ g mol}^{-1} \] Step 4: Conclusion The molar mass of solute ‘X’ is \( 57 \) g mol\(^{-1}\), so the correct answer is (B) \( 57 \) g mol\(^{-1}\).