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Chapter 13: Problem 36

Two aqueous solutions of sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11},\) are mixed. One solution is \(0.1487 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) and has \(d=1.018 \mathrm{g} / \mathrm{mL} ;\) the other is \(10.00 \% \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) by mass and has \(d=1.038 \mathrm{g} / \mathrm{mL} .\) Calculate the mole percent \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) in the mixed solution.

Short Answer

Expert verified
To get the short answer, calculate the mole percent from the total moles of sucrose and total moles in the solution using the formula from step 4.

Step by step solution

01

Amount of moles from Molarity

Use the definition of molarity to calculate the moles for the first solution. Molarity is defined as moles of solute per liter of solution. We have 0.1487 moles of sucrose (C12H22O11) per liter. We will assume 1 liter of this solution for simplicity.
02

Amount of moles from Mass Percentage

We know that the second solution is 10.00% sucrose by mass. This means that for every 100 grams of solution, 10 grams are sucrose. Calculating moles of sucrose using its molar mass (342.3 g/mol) gives us moles of sucrose in the second solution. We will assume 1 kg of this solution for simplicity.
03

Total Moles and Total Mass of the Mixed Solution

Now both solutions are mixed, so their moles of sucrose and mass of solutions should be added together to find the total moles of sucrose and total mass of the solution.
04

Calculation of Mole Percent

Mole percent is calculated as moles of the component of interest divided by the total moles in the mixed solution, multiplied by 100. For this case, it's moles of sucrose divided by total moles, times 100.

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Most popular questions from this chapter

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