Question 13.Int.1: A 50.00 g sample of a solution of naphthalene [C10H8(s)] in ...
A 50.00 g sample of a solution of naphthalene [C_{10}H_8(s)] in benzene [C_6H_6(l)] has a freezing point of 4.45 °C. Calculate the mass percent C_{10}H_8 and the boiling point of this solution.
Analyze
Equation (13.5) relates freezing-point depression to the molality of a solution, but we will have to devise an algebraic method that yields the mass of each solution component and consequently the mass percent composition. The boiling point of the solution can be determined with a minimum of calculation through equation (13.6).
\Delta T_f = -K_f \times m (13.5)
\Delta T_{ b }=K_{ b } \times m (13.6)
Substitute data for benzene from Table 13.2 (K_f = 5.12 ^{\circ}C m^{-1}, f_p = 5.53 ^{\circ}C) and the measured freezing point of the solution into equation (13.5), to obtain
| TABLE 13.2 Freezing-Point Depression and Boiling-Point Elevation Constants^a | ||||
| Solvent | Normal Freezing Point,°C | K_f, °C m^{-1} | Normal Boiling Point, °C | K_b, °C m^{-1} |
| Acetic acid | 16.6 | 3.90 | 118 | 3.07 |
| Benzene | 5.53 | 5.12 | 80.10 | 2.53 |
| Nitrobenzene | 5.7 | 8.1 | 210.8 | 5.24 |
| Phenol | 41 | 7.27 | 182 | 3.56 |
| Water | 0.00 | 1.86 | 100.0 | 0.512 |
_{}^{a}\text{Values} correspond to freezing-point depressions and boiling-point elevations, in degrees Celsius, caused by 1 mol of solute particles dissolved in 1 kg of solvent in an ideal solution.
\Delta T_{ f }=-K_{ f } \times m
(4.45-5.53) ^{\circ} C = -5.12 kg mol ^{-1} ^{\circ} C \times m
1.08 = 5.12 kg mol ^{-1} \times m
Express the molality of the solution in terms of the masses of naphthalene, x, and benzene, y, and the molar mass, 128.2 g C_{10}H_8/mol.
m = \frac{\frac{x g C _{10} H _8}{128.2 g C _{10} H _8 / mol }}{y g C _6 H _6 \times 1 kg / 1000 g C _6 H _6}
Because the total mass is 50.00 g, x + y = 50.00 and the expression above reduces to
m = \frac{\text { moles } C _{10} H _8}{ kg C _6 H _6} = \frac{(x / 128.2)}{(50.00 – x) / 1000} mol kg^{-1}
Now, substituting this expression of m into the expression derived from equation (13.5), we obtain
1.08 = 5.12 \times \frac{(x / 128.2)}{(50.00 – x) / 1000}
which we solve for x, the mass of naphthalene in grams.
\frac{1.08 \times 128.2}{5.12 \times 1000} = \frac{x}{50.00 – x}
0.0270 = \frac{x}{50.00 – x}
1.0270x = 0.0270 × 50.00
x = 1.31
The mass percent naphthalene in the solution is
\% C _{10} H _8 = \frac{1.31 g C _{10} H _8}{50.00 g \text { soln }} \times 100 \% = 2.62 \%
A simple way to find the boiling point of the solution is to first solve equation (13.5) for the molality of the solution.
\text { molality } = \frac{\Delta T_{ f }}{-K_{ f }} = \frac{-1.08 ^{\circ} C }{-5.12 ^{\circ} C m ^{-1}} = 0.211 m
Because the molality at the boiling point is the same as at the freezing point, substitute 0.211 m into equation (13.6) and solve for \Delta T_b.
\Delta T_{ b } = K_{ b } \times m = 2.53 ^{\circ} C m^{-1} \times 0.211 m = 0.534 ^{\circ} C
The boiling point of the solution is 0.534 °C higher than the normal boiling point of benzene (80.10 °C), that is,
80.10 °C + 0.53 °C = 80.63 °C
Assess
The mass of solution can easily be determined to the nearest 0.01 g and expressed with four significant figures. The freezing point of the solution can also be established to the nearest 0.01 °C with good precision. However, the freezing point depression (-1.08 °C), the difference between two numbers of comparable magnitude, is valid only to about one part per hundred. Although significant-figure rules permit three significant figures in the remainder of the calculation, the actual precision of the calculated quantities is still only about one part per hundred. The final estimate of the boiling point (80.63 °C) seems reasonably good since it required expressing \Delta T_b only to two significant figures. We assume that the 0.211 m solution is dilute enough and close enough to ideal in behavior to make equations (13.5) and (13.6) applicable.