General Chemistry Question

Question

The normal freezing point of benzene (C₆H₆) is 5.5 °C. If 29.53 grams of the nonvolatile nonelectrolyte 5-bromo-2-naphthoic acid (C₁₁H₇BrO₂), are dissolved in 170.8 grams of benzene, what is the freezing point of the resulting solution? K_fp for benzene is 5.12 °C/m.

Answer 1

We need to find the molality of the solution. Molality is defined as the moles of solute per 1000 g of solvent. We now need to calculate the number of moles of solute. The molecular weight of 5-bromo-2-naphthoic acid (C₁₁H₇BrO₂) is 251.08 g/mole; therefore, the number of moles is

29.53 g / 251.08 g/mole = 0.1176 moles. The molality of the solution then is

0.1176 moles / 170.8 g x 1000 = 0.6885 molal. The freezing point depression (Kpf is given as 5.12 C / molal, so the freezing point is depressed by 5.12 C /m x 0.6885 m = 3.525 C.

As the freezing point of pure benzene is 5.5 C, the freezing point is lowered by 3.525 C, so the freezing point is now 5.5 C – 3.525C = 1.975 C or rounding to two sig figs = 2.0 C.

Answer 2