Question

Two solutes, $0.3 \text{ gm}$ of $\text{A}$ ($M_w = 60 \text{ gm/mol}$) & $0.9 \text{ gm}$ of $\text{B}$ ($M_w = 180 \text{ gm/mol}$) are dissolved in $100 \text{ ml}$ solution. Find osmotic pressure of solution at $300 \text{ K}$ (in $\text{atm}$) ($\text{R} = 0.082 \text{ atm}-\text{L/mol-K}$)

• A. $1.23$
• B. $2.46$
• C. $4.92$
• D. $3.69$

Hint:

Osmotic pressure depends only on the total molarity of solute particles. Calculate the moles of each solute separately and sum them to find the total molar concentration $C$.

Answer 1

The Correct Option is B

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

\[\Pi = \dfrac{nRT}{V}\]

Where:

• \(\Pi\) = osmotic pressure
• \(n\) = total number of moles of solute
• \(R\) = universal gas constant = 0.082 atm-L/mol-K
• \(T\) = temperature in Kelvin = 300 K
• \(V\) = volume of solution in liters = 0.1 L (since 100 ml = 0.1 L)

We will first calculate the number of moles of each solute:

1. Solute A:

\[\text{Number of moles of A} = \dfrac{\text{mass of A}}{M_w \text{ of A}} = \dfrac{0.3 \text{ gm}}{60 \text{ gm/mol}} = 0.005 \text{ mol}\]

2. Solute B:

\[\text{Number of moles of B} = \dfrac{\text{mass of B}}{M_w \text{ of B}} = \dfrac{0.9 \text{ gm}}{180 \text{ gm/mol}} = 0.005 \text{ mol}\]

Total number of moles (\(n\)) = 0.005 mol + 0.005 mol = 0.01 mol

Substituting the values into the formula for osmotic pressure:

\[\Pi = \dfrac{nRT}{V} = \dfrac{0.01 \times 0.082 \times 300}{0.1}\]

Calculating:

\[\Pi = \dfrac{0.246}{0.1} = 2.46 \text{ atm}\]

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

The correct option is 2.46.