Maple syrup is mostly a solution of sucrose in water. Calculate the molality of the syrup if a sample freezes at -0.50^$\circ$C, assuming the solute is pure sucrose. [For water, K_f is 1.86^$\circ$C/m.] A) 0.17 m B) 0.27 m C) 0.37 m D) 047 m E) None of the above

Question

Quizplus · Accepted Answer

The freezing point depression (∆Tf) can be calculated using the formula:

∆Tf = Kf·m·i

where Kf is the freezing point depression constant for the solvent (water), m is the molality of the solution, and i is the van't Hoff factor (the number of particles into which a solute dissociates in solution). For pure sucrose, i = 1.

The freezing point of pure water is 0°C, or 32°F. The change in temperature can be calculated as:

∆Tf = (-0.50°F) - (32°F) = -32.5°F = -18.06°C

Substituting the appropriate values, we get:

-18.06°C = (1.86°C/m)·m·1

Solving for m, we get:

m = (-18.06°C) / [(1.86°C/m)·1] ≈ 0.27 m

Therefore, the molality of the syrup is approximately 0.27 m, and the best answer choice is B.

Maple Syrup Is Mostly a Solution of Sucrose in Water ∘\circ∘

Maple Syrup Is Mostly a Solution of Sucrose in Water $\circ$