Question

The osmotic pressure of solution containing 34.2 g of cane sugar (molar mass = 342 g $mo{l}^{-1}$ ) in 1L of solution at ${20}^{o}C$ is :
[Given. R = 0.082 L atm $K^{-1}mo{l}^{-1}$]

• A. 2.40 atm
• B. 3.6 atm
• C. 24 atm
• D. 0.0024 atm

Answer 1

The correct option is A 2.40 atm

The expression for the osmotic pressure of the solution is shown below.
$\pi =CRT=\frac{W}{M\times V}RT$
Here, $\pi$ is the osmotic pressure, C is the molar concentration, W is mass of solute, M is the molar mass of solute, V is the volume of the solution in liter, R is the ideal gas constant and T is the absolute temperature.
$\pi =\frac{\text{34.2 g}}{\text{342 g/mol}}\times \text{0.082 L atm / mol K}\times \text{293 K}$

$\pi =\text{2.40 atm}$
The osmotic pressure of the solution is 2.40 atm.