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Chapter 13: Problem 70

Arrange the following aqueous solutions, each \(10 \%\) by mass in solute, in order of increasing boiling point: glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right),\( sucrose \)\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),$ sodium nitrate \(\left(\mathrm{NaNO}_{3}\right)\).

Short Answer

Expert verified
The order of increasing boiling point for the given aqueous solutions with 10% by mass solute is sucrose (\(0.324 \, \text{mol/kg}\)) < glucose (\(0.617 \, \text{mol/kg}\)) < sodium nitrate (\(2.614 \, \text{mol/kg}\)).

Step by step solution

01

Calculate the moles of solute in each solution

The mass percentage of each solute in the solutions is 10% by mass. That means, for every 100 g of solution, there are 10g of solute and 90g of solvent (water). We will now calculate the moles of solute in each solution using their respective molar masses: - For glucose (\(C_{6}H_{12}O_{6}\)), Molar mass = 180 g/mol Moles of glucose = \(\frac{10 \, \text{g}}{180 \, \text{g/mol}}\)= 0.0556 moles - For sucrose (\(C_{12}H_{22}O_{11}\)), Molar mass = 342 g/mol Moles of sucrose = \(\frac{10 \, \text{g}}{342 \, \text{g/mol}}\)= 0.0292 moles - For sodium nitrate (\(NaNO_{3}\)), Molar mass = 85 g/mol Moles of sodium nitrate = \(\frac{10 \, \text{g}}{85 \, \text{g/mol}}\)= 0.1176 moles
02

Calculate the molality of each solution

Now, we will calculate the molality of each solution using the previously calculated moles of solute and the mass of solvent (water) in kilograms. - Molality of glucose = \(\frac{0.0556 \, \text{moles}}{0.090 \, \text{kg}}\)= 0.617 mol/kg - Molality of sucrose = \(\frac{0.0292 \, \text{moles}}{0.090 \, \text{kg}}\)= 0.324 mol/kg - Molality of sodium nitrate = \(\frac{0.1176 \, \text{moles}}{0.090 \, \text{kg}}\)= 1.307 mol/kg
03

Consider dissociation for sodium nitrate

Sodium nitrate is an ionic compound that dissociates in water into sodium ions (Na⁺) and nitrate ions (NO₃⁻). Thus, instead of one particle, two particles (ions) are present for every mole of sodium nitrate in solution. Therefore, we must double the molality of sodium nitrate to account for the dissociation: - Adjusted molality of sodium nitrate = \(2 \times 1.307 \, \text{mol/kg}\) = 2.614 mol/kg
04

Arrange the solutions in order of increasing boiling point

We can now arrange the solutions in order of increasing molality, as a greater number of solute particles leads to a higher boiling point: 1. Sucrose (\(0.324 \, \text{mol/kg}\)) 2. Glucose (\(0.617 \, \text{mol/kg}\)) 3. Sodium nitrate (\(2.614 \, \text{mol/kg}\)) Thus, the order of increasing boiling point is: sucrose < glucose < sodium nitrate.

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Most popular questions from this chapter

When ammonium chloride dissolves in water, the solution becomes colder. (a) Is the solution process exothermic or endothermic? (b) Why does the solution form?

Commercial aqueous nitric acid has a density of $1.12 \mathrm{~g} / \mathrm{mL}\( and is 3.7 M. Calculate the percent \)\mathrm{HNO}_{3}$ by mass in the solution.

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