When 12.1 g of the sugar sucrose (a nonelectrolyte) is dissolved in exactly 800.0 g of water, the solution has a freezing point of -0.082^$\circ$C. What is the molar mass of sucrose K_f of water is 1.86^$\circ$C/m. A) 426 g B) 99.2 g C) 178 g D) 266 g E) 343 g

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Quizplus · Accepted Answer

The freezing point depression formula is ΔTf = i * Kf * m. Since sucrose is a nonelectrolyte, i = 1. ΔTf = 0.082°C, Kf = 1.86°C/m. Solving for molality (m), m = ΔTf / Kf = 0.082 / 1.86 = 0.0441 mol/kg. The moles of sucrose = molality * kg of solvent = 0.0441 * 0.8 = 0.03528 mol. Molar mass = mass / moles = 12.1 g / 0.03528 mol = 343 g/mol.

When 121 G of the Sugar Sucrose (A Nonelectrolyte) Is Dissolved