Question

calculate the osmotic pressure at 25 degree celsius and freezing point of 1.8% aqueous solutionof glucose. Assume ideal behaviour of the solution. Take density to be 1 g/ml.

Answer 1

(i) For the freezing point, first we find ΔT_f for glucose solution using formula
ΔT_f = K_f × i × m
i for glucose is 1 since it neither dissociates nor associate in sloution
K_{f for water =} 1.86 K kg mol^-1
1.8 % solution means 1.8 g glucose is present in 100 g solution.
Thus mass of solvent = 100 - 1.8 = 98.2 g = 0.982 kg
Moles of glucose in 1.8 g = 1.8 /180 = 0.01 mol
Molality of glucose solution = 0.01 /0.982 = 0.101 m
ΔT_f = 1.86 $\times$ × 1 × $\times$ 0.101 = 0.019
T_{f , w}– T_f,_gs = 0.019
where _f , _gs ​= freezing point of 1.8 % glucose solution
273.15- T_f,_gs = 0.019
T_f, gs = 273.15- 0.019 = 273.131 K
​Thus freezing point of glucose solution is 273.131 K.

(ii) Osmotic pressure
mass of solution = 100 g
Since density = 1 g/mL
Thus volume of solution = mass/ density = 100 mL
Number of moles of glucose present = 1.8 / 180 = 0.01 mole

​Thus concentration (molarity) C of solution = 0.01 / 0.1 = 0.1 g/L

We use formula, $\pi =CRT$
π is osmotic pressure
R is gas constant = 0.082 L atm K^-1mol^-1
T is temperature in Kelvin=25 degree C =298 K
V is volume of solution in L 100cm³ = 100ml =0.1L

Thus substituting the values :
$\pi =0.1\times 0.082\times 298$
= 2.4436 atm

Thus osmotic pressure = 2.4436 atm