Question

what is the osmotic pressure of 18 % volume solution of glucose at 300k in bar unit ?​

Answer 1

Given: 18 % volume solution of glucose at 300 K.

To find: The osmotic pressure of the glucose solution.

Solution:

• The osmotic pressure of a solution at a particular temperature is the hydrostatic pressure which builds up when the solution is separated from the solvent by a semipermeable membrane and which is just sufficient to stop the phenomenon of osmosis.
• It is given by the formula,

        $\pi = \frac{w}{M} \frac{RT}{V}$

• Here, π is the osmotic pressure, w is the weight of the solute and M is the molecular mass of the solute, R is the gas constant with value 0.0821 L atm K⁻¹ mol⁻¹, T is the absolute temperature and V is the volume of the solution.
• In this case, glucose is the solute and its molecular mass is 180 g.
• The volume is 100 mL since there are 18 grams of solute per 100 mL of solvent.
• On replacing the terms with the values given in the question,

        $\pi = \frac{18}{180} \frac{0.0821 * 300K}{0.1L}$

           $= 24.63 atm$

• To find the value of osmotic pressure in bar unit, the pressure in atm unit must be multiplied by 1.01325.

        $24.63 * 1.01325 = 24.96 bar$

Therefore, the osmotic pressure of the glucose solution is 24.96 bar.