For a 5% solution of urea (Molar mass = 60 g/mol), calculate the osmotic pressure at 300 K. [R = 0·0821 L atm K^-1 mol^-1]

OR

Visha took two aqueous solutions — one containing 7·5 g of urea (Molar mass= 60 g/mol) and the other containing 42.75 g of substance Z in 100 g of water, respectively. It was observed that both the solutions were frozen at the same temperature. Calculate the molar mass of Z.

Given,
5 % urea solution means 5 g urea is present in 100 ml of solution.
moles of urea present =weight given/Molecular weight of urea
=5g / 60gmol−1=112
Hence, Concentration of urea = (moles of urea(n) /volume of solution) ×1000
= {(112)/100} ∗1000
=1012
Hence Osmotic pressure =1× (10/20) ×0.082×300 atm
=20.52atm.

OR

Given,
The mass of water (w₂) = 100 g
The mass of urea (w₁) = 7.5 g
The mass of Z (w₂) = 42.75 g
The molar mass of urea (Mm₁) = 60 g/mol
Both the solutions have same freezing point

Since both the solutions have the same freezing point, the change in freezing points (ΔT_f) for both the solutions is same.

ΔT_f = K_f*m, hence both the solutions should have the same molality.

molality = moles of solute/ weight of solvent (in Kg)

Therefore, (w₁*1000)/(Mm₁*w₂) = (w₃*1000)/(Mm*w₂) ⇒ Mm = w₃*Mm₁/w₁ = 42.75*60/7.5 = 342 g/mol Hence, The molar mass (Mm) of Z = 342 g/mol