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Chapter 12: Problem 35

A \(0.631 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution in water has a density of \(1.031 \mathrm{~g} / \mathrm{mL}\). Express the concentration of this solution as (a) mass percentage. (b) mole fraction. (c) molality.

Short Answer

Expert verified
(a) 6.00%, (b) 0.0116, (c) 0.651 m

Step by step solution

01

Determine the Molar Mass of Components

Calculate the molar mass of \( \mathrm{H}_3\mathrm{PO}_4 \). Hydrogen (H) is approximately 1 g/mol (3 atoms), phosphorus (P) is 31 g/mol, and oxygen (O) is 16 g/mol (4 atoms). Thus,\[\text{Molar mass of } \mathrm{H}_3\mathrm{PO}_4 = 3(1) + 31 + 4(16) = 98 \text{ g/mol}\]
02

Mass Percentage Calculation

Given that the molarity is \(0.631 \mathrm{M}\), which means \(0.631\) moles of \(\mathrm{H}_3\mathrm{PO}_4\) per \(1\) liter of solution. The mass of \(\mathrm{H}_3\mathrm{PO}_4\) in grams is \[0.631 \times 98 = 61.838 \text{ g}\]The total mass of solution is \[1.031 \times 1000 = 1031 \text{ g}\]Thus, the mass percent of \(\mathrm{H}_3\mathrm{PO}_4\) is \[\frac{61.838}{1031} \times 100 \approx 6.00 \%\]
03

Mole Fraction Calculation

First, calculate the moles of \(\mathrm{H}_3\mathrm{PO}_4\), which is \(0.631\) moles. The mass of the solvent (water) is \[1031 - 61.838 = 969.162 \text{ g}\]Convert this mass to moles (molar mass of water is \(18\) g/mol):\[\frac{969.162}{18} \approx 53.843 \text{ moles}\]The mole fraction is then calculated as:\[\frac{0.631}{0.631 + 53.843} \approx 0.0116\]
04

Molality Calculation

Molality is defined as moles of solute per kilogram of solvent. Using the previously calculated mass of water:\[\text{Kg of water} = \frac{969.162}{1000} \approx 0.969 \text{ kg}\]The molality is then:\[\frac{0.631}{0.969} \approx 0.651 \text{ m}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Percentage
Mass percentage is a way to express the concentration of a component in a mixture. It shows how much of the total mass is made up of that component. To find the mass percentage of a solute in a solution, you need to know two things: the mass of the solute and the total mass of the solution.
The formula for calculating mass percentage is:
  • Mass Percentage = \( \left( \frac{\text{mass of solute}}{\text{total mass of solution}} \right) \times 100 \% \)
In our exercise, we calculated the mass of the solute, \( \mathrm{H}_3\mathrm{PO}_4 \), as \( 61.838 \text{ g} \), and the total mass of the solution as \( 1031 \text{ g} \). Putting these values into the formula gave us a mass percentage of approximately \( 6.00 \% \). This tells us that \( 6.00 \% \) of the solution's mass is due to \( \mathrm{H}_3\mathrm{PO}_4 \).
Understanding mass percentage is important because it helps in comparing the concentrations of solutions. It also provides a clear, understandable proportion of the solute within the entire solution.
Mole Fraction
Mole fraction is another way to express concentration. It refers to the ratio of the number of moles of one component to the total number of moles in the mixture. This method shows the exact fraction of molecules or moles for the given component compared to others in the mixture.
To calculate the mole fraction, you need the number of moles of the solute and the number of moles of the solvent. The formula is:
  • Mole Fraction (solute) = \( \frac{\text{moles of solute}}{\text{moles of solute + moles of solvent}} \)
Using this formula in the exercise, we have \( 0.631 \) moles of \( \mathrm{H}_3\mathrm{PO}_4 \) and approximately \( 53.843 \) moles of water. Thus, the mole fraction of \( \mathrm{H}_3\mathrm{PO}_4 \) is approximately \( 0.0116 \). This result signifies that only a small fraction of the total molecules in the solution are \( \mathrm{H}_3\mathrm{PO}_4 \).
Mole fractions are essential in thermodynamic calculations and when determining chemical equilibria, as they provide a real ratio that is useful in various chemical equations.
Molality
Molality is a different concentration term that describes how many moles of solute exist per kilogram of solvent. Unlike molarity, which depends on volume, molality is based on the weight of the solvent and is thus unaffected by temperature and pressure changes.
To calculate molality, use the following formula:
  • Molality = \( \frac{\text{moles of solute}}{\text{kilograms of solvent}} \)
In the solution, the mass of the solvent (water) was calculated as \( 969.162 \text{ g} \) which is \( 0.969 \text{ kg} \). Then the number of moles of \( \mathrm{H}_3\mathrm{PO}_4 \) was \( 0.631 \text{ moles} \). Dividing these gives a molality of roughly \( 0.651 \text{ m} \). This tells us that there are \( 0.651 \) moles of \( \mathrm{H}_3\mathrm{PO}_4 \) for every kilogram of water.
Molality is particularly useful when measuring reactions or processes that involve heat, since it doesn't vary with temperature changes, making it a reliable metric in such scenarios.

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Most popular questions from this chapter

Hemoglobin contains \(0.33 \%\) Fe by mass. A \(0.200-\mathrm{g}\) sample of hemoglobin is dissolved in water to give 10.0 \(\mathrm{mL}\) of solution, which has an osmotic pressure of 5.5 torr at \(25^{\circ} \mathrm{C}\). How many moles of Fe atoms are present in 1 mol hemoglobin? (Hint: Calculate the molar mass from the osmotic pressure and find the mass of iron in one mole of the compound.)

The enthalpy of solution of nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) in water is \(-12.0 \mathrm{~kJ} / \mathrm{mol}\), and its solubility at \(20^{\circ} \mathrm{C}\) and \(2.00 \mathrm{~atm}\) is \(0.055 \mathrm{~m} .\) State whether the solubility of nitrous oxide is greater or less than \(0.055 \mathrm{~m}\) at (a) \(2.00 \mathrm{~atm}\) and \(0{ }^{\circ} \mathrm{C}\). (b) \(2.00 \mathrm{~atm}\) and \(40^{\circ} \mathrm{C}\). (c) \(1.00 \mathrm{~atm}\) and \(20^{\circ} \mathrm{C}\). (d) \(1.00 \mathrm{~atm}\) and \(50{ }^{\circ} \mathrm{C}\).

Which pairs of liquids will be soluble in each other? (a) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{6}\) (benzene) and \(\mathrm{CCl}_{4}\) (c) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

A mixture contains \(15.0 \mathrm{~g}\) hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) and \(20.0 \mathrm{~g}\) heptane \(\left(\mathrm{C}_{7} \mathrm{H}_{16}\right)\). At \(40^{\circ} \mathrm{C}\), the vapor pressure of hexane is 278 torr and that of heptane is 92.3 torr. Assume that this is an ideal solution. (a) What is the mole fraction of each of these substances in the liquid phase? (b) What are the vapor pressures of hexane and of heptane above the solution? (c) Find the mole fraction of each substance in the vapor phase.

We analyzed the enthalpy change that accompanies the formation of a solution from the pure solute and solvent by considering three changes that must occur: (1) separate the solvent molecules; (2) separate the solute molecules; and (3) bring the solute particles and solvent particles together. If the enthalpy of solution is negative (an exothermic process), what can be said about the relative enthalpy changes of the three processes just listed?
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