MP Board · Class 12 · Chemistry · Solutions19.5 g of $CH3CH2COOH$ (Molar mass = $74 \text{ g mol}^{-1}$) is dissolved in 500 g of water. Calculate the depression in freezing point and the osmotic pressure of the solution at 298 K. Given: $Kf = 1.86 \text{ K kg mol}^{-1}$, Van't Hoff factor ($i$) = 1.02, density of solution = $1 \text{ g cm}^{-3}$.

Step 1: Calculate the moles and molality of the solution

• Mass of solute ($w_2$) = $19.5 \text{ g}$
• Molar mass of solute ($M_2$) = $74 \text{ g mol}^{-1}$
• Mass of solvent ($w_1$) = $500 \text{ g} = 0.5 \text{ kg}$ \nMolality ($m$) = $\frac{w_2 \times 1000}{M_2 \times w_1} = \frac{19.5 \times 1000}{74 \times 500} = \frac{19500}{37000} = 0.527 \text{ mol kg}^{-1}$

Step 2: Calculate the depression in freezing point ($\Delta T_f$)\nUsing the formula:

$$\Delta T_f = i \cdot K_f \cdot m$$ $$\Delta T_f = 1.02 \times 1.86 \times 0.527$$ $$\Delta T_f = 1.0001 \text{ K} \approx 1.00 \text{ K}$$

Step 3: Calculate the osmotic pressure ($\pi$)\nFirst, calculate the molarity ($M$) of the solution:

• Volume of solution $\approx$ Volume of water = $500 \text{ mL} = 0.5 \text{ L}$
• Moles of solute ($n_2$) = $\frac{19.5}{74} = 0.2635 \text{ moles}$ \nMolarity ($M$) = $\frac{0.2635 \text{ moles}}{0.5 \text{ L}} = 0.527 \text{ mol L}^{-1}$ \nUsing the formula for osmotic pressure: $$\pi = i \cdot M \cdot R \cdot T$$\nWhere $R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$ and $T = 298 \text{ K}$ $$\pi = 1.02 \times 0.527 \times 0.0821 \times 298$$ $$\pi = 13.13 \text{ atm}$$