Question 7.9: Calculate the boiling and freezing points of the following s...

In each case, Equations 7.13 and 7.14 are used to calculate the difference between the normal boiling and freezing point of water and the solution. To use these equations, K_{b} and K_{f} are obtained from Table 7.6; the solution molarity, M, is calculated; and n is determined.

ΔT_{b} = nK_{b}M     (7.13)

ΔT_{f} = nK_{f}M     (7.14)

a. To find the boiling point, calculate solution molarity.

(171  \cancel{g}  \cancel{C_{12}H_{22}O_{11}} )(\frac{1  mol  C_{12}H_{22}O_{11}} {342.0  \cancel{g}   \cancel{C_{12}H_{22}O_{11}}} ) = 0.500  mol  C_{12}H_{22}O_{11}

M  = \frac{moles  of  solute}{liters  of  solution}= \frac{0.500  mol}{1.00  L} = 0.500  \frac{mol}{L}

Determine n: Because sugar does not dissociate on dissolving, n = 1. Therefore,

ΔT_{b} = nK_{b}M = (1) (0.52°C/  \cancel{M}) (0.500  \cancel{M}) = 0.26°C

Because boiling points are higher in solutions, we add ΔT_{b} to the normal boiling point of water:

solution boiling point = 100.00°C + 0.26°C = 100.26°C

To find the freezing point, calculate ΔT_{f}:

ΔT_{f} = nK_{f}M = (1) (1.86°C/\cancel{M}) (0.500  \cancel{M}) = 0.93°C

Because freezing points are lower in solutions, we subtract ΔT_{f} from the normal freez-ing point of water:

solution freezing point = 0.00°C – 0.93°C = -0.93°C

b. Similarly,

(13.4  \cancel{g}  \cancel{NH_{4}Cl}) (\frac{1  mol  NH_{4}Cl}{53.5  \cancel{g}  \cancel{NH_{4}Cl}}) = 0.250  mol NH_{4}Cl

M  = \frac{moles  of  solute}{liters  of  solution}= \frac{0.250  mol}{0.500  L} = 0.500  \frac{mol}{L}

Because NH_{4}Cl dissociates in water, 1 mol of solute gives 2 mol of particles (ions):

NH_{4}Cl(aq) → NH^{+}_{4} (aq) + Cl^{-} (aq)

Thus, we conclude that n = 2. Therefore,

ΔT_{b} = nK_{b}M = (2) (0.52°C/ \cancel{M}) (0.500  \cancel{M}) = 0.52°C

solution boiling point = 100.00°C + 0.52°C = 100.52°C

ΔT_{f} = nK_{f}M = (2) (1.86°C/\cancel{M}) (0.500  \cancel{M}) = 1.86°C

solution freezing point = 0.00°C – 1.86°C = -1.86°C