A glucose solution contains 59.0 g of glucose (C6H12O6) in 450 g of water. Part

ID: 925041 • Letter: A

Question

A glucose solution contains 59.0 g of glucose (C6H12O6) in 450 g of water.

Part A

Compute the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)

Part B

Compute the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)

Express your answer using six significant figures.

A glucose solution contains 59.0 g of glucose (C6H12O6) in 450 g of water.

Part A

Compute the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)

Part B

Compute the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)

Express your answer using six significant figures.

DTb = i*Kb*m

i = vanthoff's factor (for glucose i = 1)

m = molality

Kb = 0.512 Kg/mol

m = (wt/mol.wt)*(1000/wt of solvent in gms)

m = (59/180)*(1000/450)

m = 0.73

DTb = boiling point of solution - boiling point of water

now

(Tb -100) = 1*0.512*0.73

Tb- 100 = 0.374

Tb = 100.374

Boiling point of the glucose solution is 100.374 or 100.4 0C

.................................................................................................................................

DTf = i*Kf*m

DTf = ( freezing point of solution- freezing point of wate)

i = 1

Kf = -1.86 kg/mol

m = 0.73

now

( Tf- 0) = 1*(-1.86)*0.73

Tf - 0= -1.36

Tf = - 1.36 0C

freezing point of glucose solution is - 1.36 0C

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