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

DTf = kfm and

DTb= kbm

In the above expressions, DTf and DTb are the decrease in freezing point and increase in boiling point, respectively, and kf and kb 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 kf and kb are available in Table 13.4, as are the normal freezing and boiling points Tf and Tb 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:

Tb = 78.4 oC, kb = 1.22 oC/m; Tf = -114.6 oC, kf = 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:

convert to moles

get the molality:

for the pure solvent (CHCl3), we get from Table 13.4: Tf = -63.5oC, kf = 4.68 oC/m; Tb=61.2 oC, kb=3.63oC/m

freeze point depression:

units are (oC/m)(m)

so the solution freezes at

oC********************************************

boiling point elevation:

the solution boils at

oC***************************************************

(c) .45 mol ethylene glycol and .15 mol KBr in 150 g H2O

From table 13.4, we have for the solvent (H2O): Tf = 0oC, kf = 1.86 oC/m; Tb=100oC, kb=0.53oC/m

Yikes! we have two different solutes here. Ethylene glycol is a molecular substance

(gee, it would've been nice if you'd been given a formula so you could see that!) and KBr is ionic.

We need to get the molality of the solution: we have 0.45 mol ethylene glycol and 2(.15) mol of

ions from the KBr (why?) in .150 kg of water:

for the molality:

freeze point lowering:

(oC/m)(m)

oC

water freezes at 0oC; the solution freezes at

boiling point elevation:

(oC/m)(m)

oC

water boils at 100oC; the solution boils at

oC

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DAC 7/8/08