K b and K f depend only on the SOLVENT. Below are some common values. Use these

Question

Kb and Kf depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow.

0.512

-1.86

1.22

-1.99

3.67

2.53

-5.12

2.02

*Please note that Tf as defined above will be a negative number. Therefore, Kf must also be given as a negative number.

You may also see Tf defined as the magnitude of the freezing point depression, a positive number. In this case, Kf will also be a positive number. Regardless of the sign used for Kf, remember that the freezing point of the solution is always depressed.

Problem:

The boiling point of water/H2O is 100.00 °C at 1 atmosphere.

If 13.13 grams of calcium iodide, (293.9 g/mol), are dissolved in 263.8 grams of water ...

The molality of the solution is _____________ m.

The boiling point of the solution is ____________ °C.

T = T(solution) - T(pure solvent) * m = (# moles solute / Kg solvent) i = (# moles of solute particles / mole of solute) Kb = boiling point elevation constant. Kf = freezing point depression constant.

Explanation / Answer

Tb = Kb*mi

i = CaI2 = 1+2 = 3ions

then

m = mol S / kg solvent

mol S = mass/MW = 13.13/293.9 = 0.0446750 mol of S

kg solvnet = 0.2638 kg

molal = 0.0446750/0.2638 = 0.169351

a)

molality = 0.169351 molal

b)

Tb = Tb + dTb

dTb = 0.169351*3*0.512 = 0.2601

Tb = 100+0.2601= 100.2601 °C

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