Used to derive answers in the calculations below: a) normal boiling point of a s

Question

Used to derive answers in the calculations below: a) normal boiling point of a solution made from 18.02 g glucose dissolved in 1.250 kg of water b) normal freezing point predicted for a solution made from 58.44 g sodium chloride dissolved in 250.0 g of water c) osmotic pressure predicted for a solution made from 5.844 g sodium chloride dissolved in enough water to make 2000.0 mL of aqueous solution at 25.0 degree C d) vapor pressure above a solution made from 180.2 g glucose dissolved in 180.2 g of water at 100.0 degree C and normal pressure

Explanation / Answer

Solution:-

(a) delta Tb = i x kb x m

where  delta Tb is the elevation in boiling point.

i is the Van't Hoff factor. It's like how many ions an ionic compund give on ionization. For example i = 2 for NaCl as it gives two ions, Na+ and Cl-.

kb is the molal elevation constant and m is molality.

molality = moles of solute/kg of solvent

moles of solute = 18.02g x (1mol/180.2g) = 0.1 mol

molality = 0.1mol/1.250 kg = 0.08 m

kb for water is 0.512 0C/m

glucose is covalent molecule so, i = 1

delta Tb = 1 x 0.512 0C/m x 0.08 m

delta Tb = 0.041

Boiling point of pure water is 100 0C. So, the boiling point of the solution = 100 + 0.041 = 100.041 0C

(b) moles of solute = 58.44 g x (1mol/58.44g) = 1.00 mol

NaCl ----> Na+ + Cl-

i = 2

molality = 1.00 mol/0.250 kg = 4 m

kf = 1.82 0C/m

delta Tf = 2 x 1.82 0C/m x 4 m = 14.56 0C

pure water freezes at 0 degree C, since the depression in freezing point takes place when a on volatile solute is added to it, so the freezing point of the solution will be = -14.56 0C

(c) pi = icRT

where pi stands for osmotic pressure.

c is the concentration(molarity), R is gas constant and T is kelvin temperature.

5.844 g x (1mol/58.44 g) = 0.1 mol

as calculated in part b, i = 2

molarity = 0.1 mol/2.00 L = 0.05 M

T = 25.0 + 273 = 298 K

R = 0.0821 atm L mol-1 K-1

pi = 2 x (0.05 mol/L) x (0.0821 atm L mol-1 K-1) x 298 K

pi = 2.45 atm

(d) 180.2 g glucose x (1mol/180.2g) = 1.00 mol glucose

180.2 g water x (1mol/18.02 g) = 10.0 mol water

mole fraction of water = (moles of water)/(moles of water + moles of glucose)

mole fraction of water = 10.0/(10.0 + 1.00) = 10.0/11.0 = 0.909

vapor pressure of solution = normal vapor pressure of solvent x mole fraction

normal vapor pressure of water at 25.0 degree would be given in the book or could be checked online. At 25.0 degree C it is around 23.76 mm Hg.

So, vapor pressure of solution = 23.76 mm Hg x 0.909 = 21.60 mm Hg

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