Step 1: Understanding Isotonic Solutions Two solutions are isotonic if they possess identical osmotic pressure, calculated using the formula:\[\Pi = \frac{w}{M} \times \frac{1000}{V}\]where:- \( \Pi \) denotes osmotic pressure,- \( w \) represents the mass of solute in grams,- \( M \) signifies the molar mass of the solute,- \( V \) is the volume of the 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: A \( 6\% \) solution indicates \( 6 \) g of sucrose dissolved in \( 100 \) mL of solution.- Solute ‘X’ solution: A \( 1\% \) solution indicates \( 1 \) g of solute ‘X’ dissolved in \( 100 \) mL of solution.- Molar mass of sucrose: \( M_1 = 342 \) g mol\(^{-1}\).- Molar mass of solute ‘X’: \( M_2 \) is the value to be determined.Applying the isotonic equation:\[\frac{6}{342} = \frac{1}{M_2}\]Step 3: Solving for \( M_2 \) Rearranging the equation to solve for \( M_2 \):\[M_2 = \frac{1 \times 342}{6}\]\[M_2 = 57 \text{ g mol}^{-1}\]Step 4: Conclusion The molar mass of solute ‘X’ is calculated to be \( 57 \) g mol\(^{-1}\). Therefore, the correct answer is (B) \( 57 \) g mol\(^{-1}\).