If 1.71 g of sugar (molar mass = 342) are dissolved in 500 mL of an aqueous solution at 300 K. What will be its osmotic pressure?

Explanation

Given : Wt. of solute (w) = 1.71 g
Molar mass of solute (m) = 342
Volume of solution (V) = 500 ml = 0.5 Lt
Temperature (T) = 300K
Gas constant (R ) = 0.0821 Lt atm `K^(-1)mol^(-1)`
Osmotic pressure `(pi)=(wRT)/(mV)`
`=(1.71xx0.0821xx300)/(342xx0.5)`
= 0.25 atm
Thus, Osmotic pressure of the solution is 0.25 atm.