Question

2. You are given three solutions to test in the lab: 1.0 M glucose, 1.0 M potassium nitrate, and 1.0 M potassium phosphate. Place the solutions in order from lowest to highest

a. boiling point
b. vapor pressure
c. freezing point

Answer 1

We know that ΔT _f = iK_f x m
Where

ΔT _f = depression in freezing point

        = freezing point of pure solvent – freezing point of solution

K _f = depression in freezing constant

i= vanthoff’s factor

m = molality of the solution

So as i is more then that solution has less freezing point

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We know that ΔT _b = iK_bx m
Where

ΔT _b= elevation in boiling point

        = boiling point of solution - boilinging point of pure solvent

i= vanthoff’s factor

K_b = elevation in boiling point constant of water

m = molality of the solution

So as i is more then that solution has more boiling point

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more the number of ions ( i) less solvent molecules are converted into vapour so low vapor pressure

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Since all the solutions are having same molality the properties will depend upon the value of i

Glucose non-electrolyte so i = 1

potassium nitrate KNO3 ----> K+ + NO3 -

                                         1 +1 = 2 ions ===> i = 2

potassium phosphate K3PO4 ---> 3K+ + PO4 3-

                                                3+1 = 4ions ====> i= 4

So Order of boiling point is : Glucose < potassium nitrate < potassium phosphate

Order of freezing point : potassium phosphate < potassium nitrate < Glucose

Order of vapor pressure : potassium phosphate < potassium nitrate < Glucose