1) An aqueous solution is 13.6% nonvolatile and nonelectrolyte solute (32.05 g/m

Question

1) An aqueous solution is 13.6% nonvolatile and nonelectrolyte solute (32.05 g/mol) by mass and has a density of 0.981 g/mL. If Kf of water is 1.86 oC/m, what is the freezing point of the solution?

2) A solution contains 14.5 g of an unknown nonvolatile and nonelectrolyte solute in 100.0 g of solvent. The freezing point of the solution is 63.3 oC. If the freezing point of pure solvent is 80.6 oC and its Kf is 6.9 oC/m, what is the molar mass of the unknown?

3) The freezing point of an aqueous solution is -12.5 oC. What is its boiling point? The freezing point depression and boiling point elevation constants for water are, 1.86 oC/m and 0.52 oC/m, respectively. Boiling point of water is 100 oC.

4)

Explanation / Answer

There are multiple questions here. . i am allowed to answer only 1 at a time. I will answer question 1 for you.Please ask other as different question.
1)
Let total mass of solution be 100 gm
mass of solute = 13.6 gm
molar mass of solute = 32.05 g/mol
number of moles of solute= mass/molar mass
                   = 13.6/32.05
                    =0.42434 mol

mass of solvent = 100-13.6 = 86.4 gm = 0.0864 Kg

molality,m = number of moles of solute/mass of solvent in kG
                       =0.42434/0.0864
                      =4.9 m

use:
delta Tf= Kf *m
               = 1.86*4.9
                = 9.1 oC
So,
new freezing point = - 9.1 oC

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