Question 6.5.4: Colligative Property Calculations A solution of 5.000 g of a......
Colligative Property Calculations
A solution of 5.000 g of a solute in 100.0 g of water is heated slowly at a constant pressure of 1.00 atm and is observed to boil at 100.421°C. Estimate the molecular weight of the solute, the effective solvent vapor pressure at 25°C, and the solution freezing point at 1 atm. The necessary properties of water can be found in Table B.1.
If the values of the normal boiling point and heat of vaporization of pure water (from Table B.1) and the gas constant are substituted into Equation 6.5-4, the result is
\Delta T_{b} = T_{bs} – T_{b0} = \frac{RT^{2}_{b0}}{\Delta \hat{H}_{v}}x (6.5-4)
\Delta T_{b}(K)= \frac{[8.314 J/(mol\cdot K)](373.16 K)^{2} x}{40,656 J/mol} =28.5 x
From the measured boiling point elevation, ΔT_{b} = 0.421 K, we may deduce that the mole fraction of the solutein the solution is x = 0.421/28.5 = 0.0148. But since the solution is known to contain (5.000/M_{s}) mol of solute, where M_{s} is the solute molecular weight, and 100.0 g/18.016 g/mol = 5.551 mol of water, we may write
0.0148=(5.000 g/M_{s})/(5.000 g/M_{s} + 5.551 mol) \\ \left. \Large{\Downarrow} \right.\\ \boxed{M_{s}= 60.1 g/mol}
From Equation 6.5-2 the effective solvent vapor pressure at 25°C is determined from the vapor pressure of pure water at this temperature (found in Table B.3) as
(p^{*}_{s})_{e} = p_{s} = (1 – x) p^{*}_{s} (6.5-2)
(p^{*}_{s})_{e} = (1.000 – 0.0148) ( 23.756 mm Hg) = \boxed{23.40 mm Hg}
Finally, substituting values of the melting point and heat of fusion of water (from Table B.1) and the gas constant into Equation 6.5-5, we obtain
\Delta T_{m} = T_{m0} – T_{ms} = \frac{RT^{2}_{m0}}{\Delta \hat{H}_{m}}x (6.5-5)
\Delta T_{m} = \frac{[8.314 J/(mol\cdot K)](273.16 K)^{2} (0.0148)}{(6009.5 J/mol)} = 1.53 K = 1.53°C \\ \left. \Large{\Downarrow} \right. \\ T_{ms}= (0.000 – 1.53)°C = \boxed{- 1.53°C}