Question

Calulate the osmotic pressure at 298k and the freezing point of 1.8 percent glucose solution assuming an ideal behaviour of the slolution. Take densities to be 1g/ml and kf for water to be 1.86kw and kg mol-1(r=0 0821 1 atm.mol-1 k-1)​

Answer 1

Answer:

Osmotic pressure = 2.45 atm

Freezing point of the solution = 272.96 K

Explanation:

Given:

• Temperature = 298 K
• Density = 1 g/ml
• $\sf K_f$ for water = 1.86 K kg mol⁻¹
• R = 0.0821 L atm mol⁻¹ K⁻¹

To Find:

• Osmotic pressure of the solution
• Freezing point of the solution

Solution:

By given 1.8 g of the solute is present in 100 ml of the solution and the density is 1 g/ml

Hence,

Weight of the solvent = 100 - 1.8 = 98.2

Molar mass of the solute (glucose) = 180 g/mol

Now the osmotic pressure is given by the equation,

$\boxed{\sf \Pi=\dfrac{w_2\:R\:T}{V\:M_2}}$

where $\sf \Pi$ is the osmotic pressure, w₂ is the weight of the solute, R is the gas constant, T is the temperature, V is the volume and M₂ is the molar mass of the solute.

Here V = 100 ml = 0.1 L

Substitute the data,

$\sf \Pi=\dfrac{1.8\times 0.0821\times 298}{180\times 0.1}$

$\sf \Pi=\dfrac{44.038}{18}$

$\implies \sf 2.45\:atm$

Hence the osmotic pressure of the solvent is 2.45 atm

Now finding the freezing point.

The depression in freezing point is given by the equation,

$\boxed{\sf \Delta T_f=\dfrac{K_f\times w_2\times 1000}{M_2\times w_1} }$

where $\sf K_f$ is the Cryoscopic constant, w₂ is the weight of the solute, w₁ is the weight of the solvent and M₂ is the molar mass of the solute.

Substitute the data,

$\sf \Delta T_f=\dfrac{1.86\times 1.8\times 1000}{180\times 98.2}$

$\sf \Delta T_f=\dfrac{3348}{17676}=0.189\:K$

Now the freezing point of the solution is given by,

Freezing point of solution = Freezing point of pure solvent - Depression in freezing point

⇒ 273.15 - 0.1894

⇒ 272.96 K

Hence the freezing point of the solution is 272.96 K.