Question

Calculate the osmotic pressure of 5% solution of urea at 273 k.

(R = 0.0821 L-atm.deg^-1)

Answer 1

Since $5 \%$ solution means $\frac{5}{0.1}$ liter and mass of urea $=60 \mathrm{gmol}^{-1}$. Thus osmotic pressure,

$\pi=\frac{n \times R T}{m g} \times V=\frac{5 \times 0.0821 \times 273}{60 \times 0.1}=18.6 \mathrm{~atm}$