When a 2.850 g mixture of the sugars sucrose (C12H22O11) and fruc...

Question

When a 2.850 g mixture of the sugars sucrose (C12H22O11) and fructose (C6H12O6) was dissolved in water to a volume of 1.50 L, the resultant solution gave an osmotic pressure of 0.1843 atm at 298.0 K. What is Xsucrose of the mixture?

Accepted Answer

Step 1: Use the formula for osmotic pressure, $ \Pi = iMRT $, where $ \Pi $ is the osmotic pressure, $ i $ is the van't Hoff factor (which is 1 for non-electrolytes like sucrose and fructose), $ M $ is the molarity, $ R $ is the ideal gas constant (0.0821 L·atm/mol·K), and $ T $ is the temperature in Kelvin.. Step 2: Rearrange the formula to solve for molarity $ M $: $ M = \frac{\Pi}{iRT} $. Substitute the given values: $ \Pi = 0.1843 $ atm, $ i = 1 $, $ R = 0.0821 $ L·atm/mol·K, and $ T = 298.0 $ K.. Step 3: Calculate the total moles of solute in the solution using the molarity $ M $ and the volume of the solution (1.50 L): $ \text{moles of solute} = M \times 1.50 $ L.. Step 4: Let $ x $ be the moles of sucrose and $ y $ be the moles of fructose. The total moles of solute is $ x + y $. Use the molar masses of sucrose (342.30 g/mol) and fructose (180.16 g/mol) to set up the equation: $ 342.30x + 180.16y = 2.850 $ g.. Step 5: Solve the system of equations: $ x + y = \text{total moles of solute} $ and $ 342.30x + 180.16y = 2.850 $ g, to find the mole fraction of sucrose, $ X_{\text{sucrose}} = \frac{x}{x+y} $.