Question

An aqueous solution is 23% glucose by mass (d = 1.090 g/mL at 20.°C). Calculate its freezing point, boiling point at 1 atm, and osmotic pressure.

Answer 1

Suppose the mass of aqueous solution is 100g

so mass of glucose = 25% = 25g

Molar mass of glucose = 180g/mol

so
Moles of glucose = mass of glucose / molar mass of glucose

= 25 g / 180 g/mol = 0.139 mol

Now, mass of water = 75 g = 0.075 Kg

Now we know the formula of molality

molality = moles / mass of solvent in Kg

Molality of glucose = 0.139 mol / 0.075 Kg = 1.85 m

Now we know the formula

T = Kf   m

Where

T = Freezing point

Kf = molal freezing point depression constant 1.86 ⁰Ckg/mol

m = molality = 1.85 m

then

T = Freezing point = 1.86   1.85 m = 3.44 ⁰C

Freezing point = - 3.44 ⁰C

For Boiling point

T = Kb m

Where Kb = molal boiling point elevation constant = 0.512 ⁰C Kg/mol

So

T = 0.512 ⁰C Kg/mol 1.85 m = 0.95 ⁰C

Boiling point = BP_normal + T = 100 + 0.95 = 100.95 ⁰C

We have density = 1.090 g/mL

So Lets find volume

Density = mass / volume

So

Volume = mass density = 100 g 1.090 g/mL = 109 mL = 0.109 L

then we will calculate Molarity

Molarity = moles / Liter of solution = 0.139 mol of glucose / 0.109 L = 1.27 M

then we know the formula of osmotic pressure

Osmotic pressure = MRT

Where

M = Molarity

R = normal gas constant = 0.0821 L.atm/K

T = 20 + 273 = 293 K

Then

Osmotic Pressure = 1.27 M 0.0821 L.atm/K 293 K = 30.55 atm