15.1 Stoichiometry and composition (Page 3/5)

Siyavula textbooks: grade 10 Page 3 / 5

Molar volumes of gases

Molar volume of gases

1 mole of gas occupies

22,4 {dm}^{3}

at S.T.P.

This applies to any gas that is at standard temperature and pressure. In grade 11 you will learn more about this and the gas laws.

Molar concentrations of liquids

A typical solution is made by dissolving some solid substance in a liquid. The amount of substance that is dissolved in a given volume of liquid is known as the concentration of the liquid. Mathematically, concentration (C) is defined as moles of solute (n) per unit volume (V) of solution.

C = \frac{n}{V}

For this equation, the units for volume are $dm^{3}$ . Therefore, the unit of concentration is $mol \cdot {dm}^{- 3}$ . When concentration is expressed in $mol \cdot {dm}^{- 3}$ it is known as the molarity (M) of the solution. Molarity is the most common expression for concentration.

Do not confuse molarity (M) with molar mass (M). Look carefully at the question in which the M appears to determine whether it is concentration or molar mass.

Concentration

Concentration is a measure of the amount of solute that is dissolved in a given volume of liquid. It is measured in $mol \cdot {dm}^{- 3}$ . Another term that is used for concentration is molarity (M)

If $3,5 g$ of sodium hydroxide (NaOH) is dissolved in $2,5 {dm}^{3}$ of water, what is the concentration of the solution in $mol \cdot {dm}^{- 3}$ ?

Convert the mass of $NaOH$ into moles
$n = \frac{m}{M} = \frac{3, 5}{40} = 0, 0875 mol$
Calculate the concentration of the solution.
$C = \frac{n}{V} = \frac{0, 0875}{2, 5} = 0, 035$

The concentration of the solution is $0,035 mol \cdot {dm}^{- 3}$ or $0,035 M$

You have a $1 {dm}^{3}$ container in which to prepare a solution of potassium permanganate ( $KMnO_{4}$ ). What mass of $KMnO_{4}$ is needed to make a solution with a concentration of $0,2 M$ ?

Calculate the number of moles of $KMnO_{4}$ needed
$C = \frac{n}{V}$

therefore

$n = C \times V = 0, 2 \times 1 = 0, 2 mol$
Convert the number of moles of $KMnO_{4}$ to mass.
$m = n \times M = 0, 2 \times 158, 04 = 31, 61 g$

The mass of $KMnO_{4}$ that is needed is $31,61 g$ .

How much sodium chloride (in g) will one need to prepare $500 {cm}^{3}$ of solution with a concentration of $0,01 M$ ?

Convert all quantities into the correct units for this equation
$V = \frac{500}{1 000} = 0, 5 {dm}^{3}$
Calculate the number of moles of sodium chloride needed
$n = C \times V = 0, 01 \times 0, 5 = 0, 005 mol$
Convert moles of sodium chloride to mass
$m = n \times M = 0, 005 \times 58, 45 = 0, 29 g$

The mass of sodium chloride needed is $0,29 g$

Molarity and the concentration of solutions

$5,95 g$ of potassium bromide was dissolved in $400 {cm}^{3}$ of water. Calculate its molarity.
100 g of sodium chloride (NaCl) is dissolved in 450 cm 3 of water.
1. How many moles of NaCl are present in solution?
2. What is the volume of water (in ${dm}^{3}$ )?
3. Calculate the concentration of the solution.
4. What mass of sodium chloride would need to be added for the concentration to become $5,7 mol \cdot {dm}^{- 3}$ ?
What is the molarity of the solution formed by dissolving $80 g$ of sodium hydroxide ( $NaOH$ ) in $500 {cm}^{3}$ of water?
What mass (g) of hydrogen chloride ( $HCl$ ) is needed to make up $1000 {cm}^{3}$ of a solution of concentration $1 mol \cdot {dm}^{- 3}$ ?
How many moles of $H_{2} SO_{4}$ are there in $250 {cm}^{3}$ of a $0,8 M$ sulphuric acid solution? What mass of acid is in this solution?

Stoichiometric calculations

Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is also the numerical relationship between reactants and products. In representing chemical change showed how to write balanced chemical equations. By knowing the ratios of substances in a reaction, it is possible to use stoichiometry to calculate the amount of either reactants or products that are involved in the reaction. The examples shown below will make this concept clearer.

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Siyavula textbooks: grade 10 physical science [caps]. OpenStax CNX. Sep 30, 2011 Download for free at http://cnx.org/content/col11305/1.7

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Siyavula textbooks: grade 10 physical science [caps]' conversation and receive update notifications?

Ask

	3 Neuroscience Exam 2004 4 By David Corey Start Exam
©flickr:	The Minute Man Hose Load By Michael Pitt Start Quiz
	25 Toxicology Dr. Gustafson/Hamar quiz By Brooke Delaney Start Exam
	1 AP 01 Human Body Anatomy Physiology Essay By OpenStax Start Flashcards
	15 Biology 15 Genes and Proteins MCQ By OpenStax Start Quiz
	Psychology By OpenStax Read Online Course
	1 Sociology 01 An Introduction to Sociology MCQ By OpenStax Start Quiz
	Does your crush like you? By Hoy Wen Start Quiz
	5 AP 05 Integumentary System MCQ By OpenStax Start Quiz
	15 AP Key Terms 15 Autonomic Nervous System By OpenStax Start Key Terms