Question

A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute A of molar mass 60 g mol$^{-1}$ and 0.9 g of a non-volatile non-electrolyte solute B of molar mass 180 g mol$^{-1}$ in 100 mL H$_2$O at 27$^\circ$C. Osmotic pressure of the solution will be
[Given: R = 0.082 L atm K$^{-1}$ mol$^{-1}$]

• A. 1.23 atm
• B. 0.82 atm
• C. 2.46 atm
• D. 1.47 atm

Hint:

For non-electrolytes, osmotic pressure depends only on the total number of solute particles present in the solution.

Answer 1

The Correct Option is C

To calculate the osmotic pressure of the solution, we will use the formula for osmotic pressure:

\[\Pi = \dfrac{n_{\text{total}} \cdot R \cdot T}{V}\]

Where:

• \(\Pi\) = osmotic pressure
• \(n_{\text{total}}\) = total number of moles of solutes
• \(R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}\)
• \(T = \text{Temperature in Kelvin} = 27^\circ\text{C} = 300 \, \text{K}\)
• \(V\) = Volume of solution in liters

First, let's calculate the number of moles of each solute.

1. Calculate moles of solute A:

Molar mass of solute A = 60 g/mol

Mass of solute A = 0.3 g

Using the formula \(n = \dfrac{\text{mass}}{\text{molar mass}}\),

\(n_A = \dfrac{0.3}{60} = 0.005 \, \text{mol}\)

2. Calculate moles of solute B:

Molar mass of solute B = 180 g/mol

Mass of solute B = 0.9 g

Using the formula \(n = \dfrac{\text{mass}}{\text{molar mass}}\),

\(n_B = \dfrac{0.9}{180} = 0.005 \, \text{mol}\)

3. Calculate the total number of moles:

Total moles, \(n_{\text{total}} = n_A + n_B = 0.005 + 0.005 = 0.01 \, \text{mol}\)

4. Convert the volume of the solution to liters:

\(V = 100 \, \text{mL} = 0.1 \, \text{L}\)

5. Calculate osmotic pressure:

Substitute all values into the osmotic pressure formula:

\[\Pi = \dfrac{0.01 \cdot 0.082 \cdot 300}{0.1}\]\[\Pi = \dfrac{0.246}{0.1} = 2.46 \, \text{atm}\]

Therefore, the osmotic pressure of the solution is 2.46 atm.

The correct answer is 2.46 atm.