A linear graph is plotted with osmotic pressure ( in atm) on the y-axis against the solute concentra — Solutions and Colligative Properties Chemistry Question
Question
A linear graph is plotted with osmotic pressure ($\pi$ in atm) on the y-axis against the solute concentration ($C$ in $\text{g cm}^{-3}$) on the x-axis for a dilute aqueous solution at exactly $27^\circ\text{C}$. If the measured slope of this straight line is exactly $4.92\text{ atm cm}^3\text{ g}^{-1}$, determine the exact molar mass of the dissolved solute in $\text{g mol}^{-1}$. (Use $R = 0.082\text{ L atm K}^{-1}\text{ mol}^{-1}$).
💡 Solution & Explanation
The osmotic pressure equation is $\pi = \frac{n}{V(\text{in L})} RT$. We can rewrite this using mass ($w$): $\pi = \frac{w}{M \times V(\text{in L})} RT$. Given concentration $C$ is in $\text{g cm}^{-3}$, so $C = \frac{w}{V(\text{in mL})}$. Thus, $V(\text{in L}) = \frac{V(\text{in mL})}{1000}$. Substituting this gives $\pi = C \times \frac{1000 \times RT}{M}$. The graph of $\pi$ vs $C$ is a straight line passing through the origin with $\text{slope} = \frac{1000 \times RT}{M} = 4.92$. Temperature $T = 300\text{ K}$, so $RT = 0.082 \times 300 = 24.6$. Solving for $M$: $M = \frac{1000 \times 24.6}{4.92} = \frac{24600}{4.92} = 5000\text{ g mol}^{-1}$.