Question

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?

Answer 1

Given: Mass of sugar = 1.71 g

Molar mass of sugar = 342 g

Volume of solution = 500 ml = 0.5 L

Temperature = 300 K

Gas constant R = 0.0821 L atm mol⁻¹ K⁻¹

Formula:

Number of moles = `"Mass of sugar"/"Molar mass"`

Molarity = `"Number of moles of solute"/"Volume of the solution in litres"`

Osmotic pressure ⇒ π = CRT

Calculation: 

Number of moles = `1.71/342`

= 0.005 mol

Molarity (C) = `(0.005 xx 1000)/500`

= 0.01 M

Osmotic Pressure = CRT

= 0.01 × 0.0821 × 300

= 0.2463 bar