Question

An aqueous solution of glucose boils at 100.02°C. What is the number of glucose molecules in the solution containing 100 g of water? What will be the osmotic pressure of this glucose solution at 27°C?

(Given: K_b for water = 0.5 K kg mol⁻¹, R = 0.082 L atm mol⁻¹ K⁻¹, Avogadro’s constant, N_A = 6.02 × 10²³ mol⁻¹)

Answer 1

Given: Boiling point elevation (ΔT_b) = 100.02°C − 100°C = 0.02 K

K_b (water) = 0.5 K kg/mol

Mass of solvent (water) = 100 g = 0.1 kg

R = 0.082 L atm mol⁻¹K⁻¹

T = 27°C = 300 K

N_A = 6.02 × 10²³ mol⁻¹

ΔT_b = K_b . m

⇒ `m = (Delta T_b)/(K_b)`

= `0.02/0.5`

m = 0.04 mol/kg

Since water = 0.1 kg,

Moles of glucose (n) = 0.04 × 0.1

= 0.004 mol

Number of molecules = n × N_A

​= 0.004 × 6.02 × 10²³

= 2.408 × 10²³ molecules

`pi = n/V RT`

Assume the solution volume ≈ volume of water = 100 mL = 0.1 L

`pi = 0.004/0.1 xx 0.082 xx 300`

= 0.04 × 0.082 × 300

= 0.984 atm

∴ The number of glucose molecules is 2.408 × 10²³ and osmotic pressure is 0.984 atm.