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

Osmotic pressure $\pi$ is calculated using $\pi = C_{\text{total}} R T$. (Assuming $i=1$).
Total concentration $C_{\text{total}} = C_A + C_B$. Volume $V = 0.1 \text{ L}$.
$C_A = \frac{n_A}{V} = \frac{0.3/60 \text{ mol}}{0.1 \text{ L}} = \frac{0.005}{0.1} = 0.05 \text{ M}$.
$C_B = \frac{n_B}{V} = \frac{0.9/180 \text{ mol}}{0.1 \text{ L}} = \frac{0.005}{0.1} = 0.05 \text{ M}$.
$C_{\text{total}} = 0.05 + 0.05 = 0.10 \text{ M}$.
$\pi = (0.10 \text{ mol/L}) \times (0.082 \text{ atm}-\text{L/mol-K}) \times (300 \text{ K})$.
$\pi = 0.1 \times 24.6 = 2.46 \text{ atm}$.