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Chapter 12: Problem 38

a. Suppose you wanted to dissolve 294.3 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in 1.000 \(\mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{O}\) . (1) What is the solute? (2) What is the solvent? (3) What is the molality of this solution? b. What is the molality of a solution of 63.0 g HNO \(_{3}\) in 0.250 \(\mathrm{kg} \mathrm{H}_{2} \mathrm{O} ?\)

Short Answer

Expert verified
The solute is \( \text{H}_{2}\text{SO}_{4}\text{ } \) and the solvent is 1.000 \( \text{ kg of } \text{H}_{2}\text{O} \. The molality of the \( \text{H}_{2}\text{SO}_{4}\text{ } \) solution is 3.00 \( \text{ m} \. The molality of the \( \text{HNO}_{3}\text{ } \) solution is 4.00 \( \text{ m} \.

Step by step solution

01

Identify Solute and Solvent

For part (1) and (2) of the exercise, recognize that the solute is the substance being dissolved and the solvent is the substance doing the dissolving. In this case, \(\text{H}_{2}\text{SO}_{4}\) is the solute and \(1.000 \text{ kg of }\text{H}_{2}\text{O}\) is the solvent.
02

Calculate Moles of \( \text{H}_{2}\text{SO}_{4} \)

First, determine the molar mass of \(\text{H}_{2}\text{SO}_{4}\). The molecular weights are: H = 1.01 g/mol, S = 32.07 g/mol, O = 16.00 g/mol. So, \(\text{H}_{2}\text{SO}_{4}\) has a molar mass of \((2 \times 1.01) + 32.07 + (4 \times 16.00) = 98.09 \text{ g/mol}\). Next, calculate the moles of \(\text{H}_{2}\text{SO}_{4}\) using the given mass: \[ \frac{294.3 \text{ g }}{98.09 \text{ g/mol }} \text{ = } 3.00 \text{ mol} \]
03

Calculate Molality of \(\text{H}_{2}\text{SO}_{4}\) Solution

Molality (m) is calculated using the formula: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg} } \] Here: \frac {3.00 \text{ mol}}{1.000 \text{ kg}} = 3.00 \text{ m } .
04

Calculate Moles of \( \text{HNO}_{3} \)

Determine the molar mass of \(\text{HNO}_{3}\). The molecular weights are: H = 1.01 g/mol, N = 14.01 g/mol, O = 16.00 g/mol. So, \( \text{HNO}_{3} \) has a molar mass of \((1 \times 1.01) + 14.01 + (3 \times 16.00) = 63.01 \text{ g/mol}\). Next, calculate the moles of \( \text{HNO}_{3}\text{ } \) using the given mass: \ \frac{63.0 \text{ g}}{63.01 \text{ g/mol }} \text{ = 1.00}\text{ mol} \.
05

Calculate Molality of \( \text{HNO}_{3}\text{ } \) Solution

Molality (m) is calculated using the formula: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg } } \] Here: \frac{ 1.00 \text{ mol }}{0.250 \text{ kg }} = 4.00 \text{ m}\ .

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Key Concepts

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

Solute
The solute is the substance being dissolved in a solution. In simpler terms, it is the ingredient that gets mixed into a solvent to form a solution. For example, if you dissolve sugar in water, sugar is the solute. Solutes can be solid, liquid, or gas:
  • Solids: Like salt or sugar.
  • Liquids: Like alcohols.
  • Gases: Like carbon dioxide in fizzy drinks.
In the provided exercise, \(\text{H}_{2}\text{SO}_{4}\) and \(\text{HNO}_{3}\) are the solutes. They are being dissolved in water (the solvent). Understanding this helps us know what we're mixing in a solution.
Solvent
The solvent is the substance that dissolves the solute. It is usually the component present in the greater amount. Most solutions we're familiar with involve water as the solvent, which is why water is often called the 'universal solvent.' Examples of solvents include:
  • Water, making up the majority of the solution in our exercise.
  • Alcohol, used in disinfectants and tinctures.
  • Acetone, used in nail polish removers.
In the provided exercise, the solvent is water (\(\text{H}_{2}\text{O}\)). The water does the dissolving action on \(\text{H}_{2}\text{SO}_{4}\) and \(\text{HNO}_{3}\).
Solution Concentration
Solution concentration explains how much solute is present in a given amount of solvent. Different ways to express concentration include:
  • Molarity (M): Moles of solute per liter of solution.
  • Molality (m): Moles of solute per kilogram of solvent, which we use in this exercise.
  • Mass percent: Mass of solute per mass of solution, multiplied by 100.
To calculate molality, divide the moles of solute by the kilograms of solvent. This helps us understand the solution's strength or dilution. For example, in our exercise, 3.00 mol of \(\text{H}_{2}\text{SO}_{4}\) in 1.000 kg of water gives a molality of 3.00 m. Similarly, 1.00 mol of \(\text{HNO}_{3}\) in 0.250 kg of water gives a molality of 4.00 m.
Molar Mass
Molar mass is the mass of one mole of a substance (typically measured in g/mol). It is the sum of the atomic masses of all atoms in a molecule. Calculating molar mass helps us convert mass into moles, which is crucial for finding solution concentration. For example, to find the molar mass of \(\text{H}_{2}\text{SO}_{4}\):
\((2 \times 1.01) + 32.07 + (4 \times 16.00) = 98.09 \text{ g/mol}\)

For \(\text{HNO}_{3}\):
\((1 \times 1.01) + 14.01 + (3 \times 16.00) = 63.01 \text{ g/mol}\)

Using molar mass, we convert mass to moles. For example, 294.3 g of \(\text{H}_{2}\text{SO}_{4}\) divided by its molar mass (98.09 g/mol) gives 3.00 mol.

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Most popular questions from this chapter

The enthalpy of solution for AgNO \(_{3}\) is \(+22.8 \mathrm{kJ} / \mathrm{mol}\) . a. Write the equation that represents the dissolution of AgNO \(_{3}\) in water. b. Is the dissolution process endothermic or exothermic? Is the crystallization process endothermic or exothermic? c. As AgNO \(_{3}\) dissolves, what change occurs in the temperature of the solution? d. When the system is at equilibrium, how do the rates of dissolution and crystallization compare? e. If the solution is then heated, how will the rates of dissolution and crystallization be affected? Why? f. How will the increased temperature affect the amount of solute that can be dissolved? g. If the solution is allowed to reach equilibrium and is then cooled, how will the system be affected?

A solution is made by dissolving 26.42 g of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in enough \(\mathrm{H}_{2} \mathrm{O}\) to make 50.00 \(\mathrm{mL}\) of solution. a. What is the molar mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) ? b. What is the molarity of this solution?

a. What rule of thumb is useful for predicting whether one substance will dissolve in another? b. Describe what the rule means in terms of various combinations of polar and nonpolar solutes and solvents.

a. Suppose you wanted to dissolve 106 \(\mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in enough \(\mathrm{H}_{2} \mathrm{O}\) to make 6.00 \(\mathrm{L}\) of solution. (1) What is the molar mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) (2) What is the molarity of this solution? b. What is the molarity of a solution of 14.0 \(\mathrm{g} \mathrm{NH}_{4} \mathrm{Br}\) in enough \(\mathrm{H}_{2} \mathrm{O}\) to make 150 \(\mathrm{mL}\) of solution?

75.0 \(\mathrm{mL}\) of an AgNO \(_{3}\) solution reacts with enough Cu to produce 0.250 \(\mathrm{g}\) of Ag by single displacement. What is the molarity of the initial AgNO \(_{3}\) solution if \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is the other product?
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