Question 11.11: USING BOILING-POINT ELEVATION TO CALCULATE THE MOLALITY OF A...

USING BOILING-POINT ELEVATION TO CALCULATE THE MOLALITY OF A SOLUTION

What is the molality of an aqueous glucose solution if the boiling point of the solution at 1 atm pressure is 101.27 °C? The molal boiling-point-elevation constant for water is given in Table 11.4.

TABLE 11.4 Molal Boiling-Point-Elevation Constants (K_{b}) and Molal Freezing-Point-Depression Constants (K_{f}) for Some Common Substances

Substance                    K_{b}[(°C · kg)/mol]            K_{f}[(°C · kg)/mol]
Benzene (C_{6}H_{6})                               2.64                                5.07
Camphor (C_{10}H_{16}O)                       5.95                               37.8
Chloroform (CHCl_{3})                        3.63                               4.70
Diethyl ether (C_{4}H_{10}O)                  2.02                               1.79
Ethyl alcohol (C_{2}H_{6}O)                   1.22                                1.99
Water (H_{2}O)                                     0.51                                1.86

STRATEGY
Rearrange the equation for molal boiling-point elevation to solve for m:

ΔT_{b}  =  K_{b}  ·  m               so                m  =  \frac{ΔT_{b}}{K_{b}}

where K_{b} = 0.51 (°C · kg)/mol and ΔT_{b} = 101.27 °C – 100.00 °C = 1.27 °C.

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m = \frac{1.27  °C}{0.51\frac{°C  ·  kg}{mol}}  =  2.5 \frac{mol}{kg} = 2.5 m

The molality of the solution is 2.5 m.

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