If a 25.0-g sample of a nonelectrolyte is dissolved in 145.0 g of water,the resulting solution will freeze at -0.89°C.What is the molar mass of the nonelectrolyte? (K_fp for water is 1.858°C/m.) A) 76 g/mol B) 0.36 g/mol C) 990 g/mol D) 290 g/mol E) 360 g/mol

Question

Quizplus · Accepted Answer

First, we need to calculate the change in freezing point of water:
ΔT = Kf · m
where ΔT is the change in temperature (in °C), Kf is the freezing point depression constant for water (1.858 °C/m), and m is the molality of the solution (in mol/kg). We can solve for molality:
m = ΔT / Kf = (-0.89 °C) / (1.858 °C/m) = -0.478 m

Since we have 25.0 g of the nonelectrolyte and 145.0 g of water, we can find the number of moles of the nonelectrolyte:
n = m / kg solvent = m / (145.0 g / 1000) = -3.3 mol

The negative sign indicates that we made an error in setting up the problem (perhaps the mass of the nonelectrolyte was supposed to be subtracted from the mass of the solution, rather than added?). We need to take the absolute value of n to find the moles of solute:
n = 3.3 mol

Finally, we can calculate the molar mass of the nonelectrolyte:
M = m / n = (25.0 g) / (3.3 mol) = 7.58 g/mol

Therefore, the best choice is E, 360 g/mol.

If a 250-G Sample of a Nonelectrolyte Is Dissolved in 145