Chemistry Formula

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ΔT1 = Kfm
where Kf is a constant that depends on the specific solvent and m is the molality of the molecules or ions solute. Table 1 gives data for several common solvents. Table 1. Molal Freezing Point and Boiling Point Constants| Solvent| Formula| Freezing Point (°C)| Kf(°C/molal)| Boiling Point (°C)| Kb(°C/molal)| Water| H2O| 0.0| 1.86| 100.0| 0.51|

Acetic acid| CH3COOH| 17.0| 3.90| 118.1| 3.07|
Benzene| C6H6| 5.5| 4.90| 80.2| 2.53|
Chloroform| CHCl3| –63.5| 4.68| 61.2| 3.63|
Ethanol| C2H5OH| –114.7| 1.99| 78.4| 1.22|
Phenol| C6H5OH| 43.0| 7.40| 181.0| 3.56|
Use the previous formula and the constant from Table 1 to calculate the temperature at which a solution of 50 grams of sucrose (C12H22O11) in 400 grams of water will freeze. The molecular weight of sucrose is 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mole

so, the number of moles of sucrose is

and the concentration of the solution in moles per kilogram of water is

By taking the freezing point constant for water as 1.86 from Table and then substituting the values into the equation for freezing point depression, you obtain the change in freezing temperature: Δ Tf = 1.86°C/m × 0.365 m = 0.68°C

Because the freezing point of pure water is 0°C, the sucrose solution freezes at –0.68°C. A similar property of solutions is boiling point elevation. A solution boils at a slightly higher temperature than the pure solvent. The change in the boiling point is calculated from Δ Tb = Kb m

where Kb is the molal boiling point constant and m is the concentration of the solute expressed as molality. The boiling point data for some solvents are provided in Table 1. Notice that the change in freezing or boiling temperature depends solely on thenature of the solvent, not on the identity of the solute. One valuable use of these relationships is to determine the molecular mass of various dissolved substances. As an example, perform such a calculation to find the...
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