Osmotic pressure exerted by a solution prepared by dissolving one gram mole of a solute in 22.4 litres of a solvent at 0°C will be __________ atmosphere.

ANS:B - 1

To calculate the osmotic pressure (ΠΠ) of a solution, you can use the formula: Π=𝑛𝑉×𝑅×𝑇Π=Vn​×R×T Where:

• 𝑛n is the number of moles of solute
• 𝑉V is the volume of the solution in liters
• 𝑅R is the ideal gas constant (0.0821 L·atm/mol·K)
• 𝑇T is the temperature in Kelvin

Given that one gram mole of a solute is dissolved in 22.4 liters of a solvent at 0°C (273.15 K), and we're assuming ideal behavior, we can substitute these values into the formula: Π=122.4×0.0821×273.15Π=22.41​×0.0821×273.15 Π=0.0821×273.1522.4Π=22.40.0821×273.15​ Π≈1 atmΠ≈1 atm So, the osmotic pressure exerted by the solution is approximately 1 atmosphere. Thus, the correct answer is 1.