Ch 202 Problem 13.70

By using data from Table 13.4, calculate the freezing and boiling points for each of the following solutions:

(a) 0.30 m glucose in ethanol

In these problems, we are concerned with calculating the freezing point lowering and boiling point elevation due to the addition of a nonvolatile solute to a given solvent. The pertinent equations are

DT_f = k_fm and

DT_b= k_bm

In the above expressions, DT_f and DT_b are the decrease in freezing point and increase in boiling point, respectively, and k_f and k_b are the molal freezing point depression constant and the molal boiling point elevation constants, respectively (units of oC/m). m is the molality of the solution.

The constants k_f and k_b are available in Table 13.4, as are the normal freezing and boiling points T_f and T_b of the pure solvents. Pay attention to the solute type (nonelectrolyte or strong or weak electrolyte) - we are dealing with colligative properties, which depend only on the number of solute particles and not on their nature.

for part (a), notice that glucose is a molecular substance; hence, we can use the molality as given in the problem. for the solvent (ethanol), we have:

T_b = 78.4 ^oC, k_b = 1.22 ^oC/m; T_f= -114.6 ^oC, k_f= 1.99 ^oC/m.

find the change in the freeze point:

units are (^oC/m)(m)

^oC - this is how much the freeze point decreases

so the freeze point of the solution is

pure ethanol freezes at -114.6; the solution freezes at -115.

similarly, for the boiling point elevation,

units are (^oC/m)(m)

^oC - the boiling point increases by this amount

now, pure ethanol boils at 78.4oC; the solution boils at

or 79 ^oC***********************************

(b) 20.0 g of decane, C₁₀H₂₂, in 45.5 g of CHCl₃

Here,we have to find the molality of the solution. The solute (C₁₀H₂₂) is a nonelectrolyte (it's a molecular substance), so we convert grams to moles:

the molar mass: