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Chapter 12: Problem 19

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?

Short Answer

Expert verified
Molar mass of \(\text{Na}_{2}\text{CO}_{3}\) is 106.0 g/mol, molarity is 0.167 M. Molarity of \(\text{NH}_{4}\text{Br}\) solution is 0.953 M.

Step by step solution

01

Calculate the Molar Mass of Na2CO3

To find the molar mass of \(\text{Na}_{2}\text{CO}_{3}\), sum up the atomic masses of all the atoms in the compound:- Sodium (Na): 2 atoms, each with an atomic mass of 23.0 g/mol- Carbon (C): 1 atom, with an atomic mass of 12.0 g/mol- Oxygen (O): 3 atoms, each with an atomic mass of 16.0 g/molThus, the molar mass is \(\text{(2*23.0) + 12.0 + (3*16.0) = 106.0}\) g/mol.
02

Calculate the Molarity of the Na2CO3 Solution

Molarity (\text{M}) is defined as the number of moles of solute per liter of solution. First, find the number of moles of \(\text{Na}_{2}\text{CO}_{3}\): \(\text{Moles of } \text{Na}_{2}\text{CO}_{3} = \frac{106 \text{ g}}{106.0 \text{ g/mol}} = 1 \text{ mol}\). Then, since the volume of the solution is 6.00 L, the molarity of the solution is: \(\text{M} = \frac{1 \text{ mol}}{6.00 \text{ L}} = 0.167 \text{ M}\).
03

Calculate the Molar Mass of NH4Br

To find the molar mass of \(\text{NH}_{4}\text{Br}\), sum up the atomic masses of all the atoms in the compound:- Nitrogen (N): 1 atom, with an atomic mass of 14.0 g/mol- Hydrogen (H): 4 atoms, each with an atomic mass of 1.0 g/mol- Bromine (Br): 1 atom, with an atomic mass of 80.0 g/molThus, the molar mass is \(\text{14.0 + (4*1.0) + 80.0 = 98.0}\) g/mol.
04

Calculate the Molarity of the NH4Br Solution

Molarity (\text{M}) is defined as the number of moles of solute per liter of solution. First, find the number of moles of \(\text{NH}_{4}\text{Br}\): \(\text{Moles of } \text{NH}_{4}\text{Br} = \frac{14.0 \text{ g}}{98.0 \text{ g/mol}} = 0.143 \text{ mol}\). Then, since the volume of the solution is 150 mL (which is 0.150 L), the molarity of the solution is: \(\text{M} = \frac{0.143 \text{ mol}}{0.150 \text{ L}} = 0.953 \text{ M}\).

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Key Concepts

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

Molar Mass
Understanding molar mass is crucial in solution chemistry. It is the mass of one mole of a given substance. To calculate the molar mass, sum up the atomic masses of all atoms in a molecule.
For example, to find the molar mass of \(\text{Na}_2\text{CO}_3\), you need: \[\text{2} \times \text{Na (23.0 g/mol)} + \text{C (12.0 g/mol)} + \text{3} \times \text{O (16.0 g/mol)} = 106.0 \text{ g/mol}\]
The molar mass tells us how much one mole of \(\text{Na}_2\text{CO}_3\) weighs. Similarly, you can calculate the molar mass of any compound by adding up the masses of its constituent atoms.
Solution Chemistry
Solution chemistry focuses on the preparation and properties of solutions. A solution consists of a solute (substance being dissolved) and a solvent (substance doing the dissolving).
Molarity (\text{M}) is a common way to express the concentration of a solution, defined as moles of solute per liter of solution. In our exercise, to make a solution from 106 g of \(\text{Na}_2\text{CO}_3\) in 6.00 L of water, you first find the number of moles, which is 1 mole, because 106 g/106 g/mol = 1 mole.
The molarity then is \(\text{M} = \frac{1 \text{ mol}}{6.00 \text{ L}} = 0.167 \text{ M}\). This shows that the solution has a concentration of 0.167 M.
Moles
A mole is a fundamental unit in chemistry that measures the amount of substance. One mole corresponds to Avogadro's number, which is approximately \( 6.022 \times 10^{23}\) entities (like atoms or molecules).
Using moles simplifies reactions and calculations significantly. For instance, in our exercise with \( \text{NH}_4\text{Br}\), you have 14.0 g of the substance. First, calculate how many moles this is: \(\frac{14.0 \text{ g}}{98.0 \text{ g/mol}} = 0.143 \text{ mol}\).
Moles help convert between the mass of a substance and its chemical behavior in reactions. Understanding this concept allows accurate calculations of reactant and product quantities in chemical reactions.

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Most popular questions from this chapter

How many milliters of 0.750 \(\mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) are required to react with 250 \(\mathrm{mL}\) of 0.150 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) if the products are barium phosphate and water?

Citric acid is one component of some soft drinks. Suppose that 2.00 L of solution are made from \(150 .\) mg of citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) a. What is the molar mass of citric acid? b. What is the molarity of citric acid in the solution?

What is the difference between an electrolyte and a nonelectrolyte?

a. Suppose you wanted to produce 1.00 \(\mathrm{L}\) of a 3.50 \(\mathrm{M}\) aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (1) What is the solute? (2) What is the solvent? (3) How many grams of solute are needed to make this solution? b. How many grams of solute are needed to make 2.50 L of a 1.75 \(\mathrm{M}\) solution of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} ?\)

In what units is molarity expressed?
See all solutions

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