# 15.1 Stoichiometry and composition  (Page 3/5)

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## Molar volumes of gases

Molar volume of gases
1 mole of gas occupies $22,4{\mathrm{dm}}^{3}$ at S.T.P.

## 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 $\mathrm{dm}{}^{3}$ . Therefore, the unit of concentration is $\mathrm{mol}·{\mathrm{dm}}^{-3}$ . When concentration is expressed in $\mathrm{mol}·{\mathrm{dm}}^{-3}$ 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 $\mathrm{mol}·{\mathrm{dm}}^{-3}$ . Another term that is used for concentration is molarity (M)

If $3,5\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of sodium hydroxide (NaOH) is dissolved in $2,5\phantom{\rule{2pt}{0ex}}{\mathrm{dm}}^{3}$ of water, what is the concentration of the solution in $\mathrm{mol}·{\mathrm{dm}}^{-3}$ ?

1. $n=\frac{m}{M}=\frac{3,5}{40}=0,0875\phantom{\rule{2pt}{0ex}}\mathrm{mol}$
2. $C=\frac{n}{V}=\frac{0,0875}{2,5}=0,035$

The concentration of the solution is $0,035\phantom{\rule{2pt}{0ex}}\mathrm{mol}·{\mathrm{dm}}^{-3}$ or $0,035\phantom{\rule{2pt}{0ex}}\mathrm{M}$

You have a $1\phantom{\rule{2pt}{0ex}}{\mathrm{dm}}^{3}$ container in which to prepare a solution of potassium permanganate ( $\mathrm{KMnO}{}_{4}$ ). What mass of $\mathrm{KMnO}{}_{4}$ is needed to make a solution with a concentration of $0,2\phantom{\rule{2pt}{0ex}}\mathrm{M}$ ?

1. $C=\frac{n}{V}$

therefore

$n=C×V=0,2×1=0,2\phantom{\rule{2pt}{0ex}}\mathrm{mol}$
2. $m=n×M=0,2×158,04=31,61\phantom{\rule{2pt}{0ex}}\mathrm{g}$

The mass of $\mathrm{KMnO}{}_{4}$ that is needed is $31,61\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .

How much sodium chloride (in g) will one need to prepare $500\phantom{\rule{2pt}{0ex}}{\mathrm{cm}}^{3}$ of solution with a concentration of $0,01\phantom{\rule{2pt}{0ex}}\mathrm{M}$ ?

1. $V=\frac{500}{1 000}=0,5\phantom{\rule{2pt}{0ex}}{\mathrm{dm}}^{3}$
2. $n=C×V=0,01×0,5=0,005\phantom{\rule{2pt}{0ex}}\mathrm{mol}$
3. $m=n×M=0,005×58,45=0,29\phantom{\rule{2pt}{0ex}}\mathrm{g}$

The mass of sodium chloride needed is $0,29\phantom{\rule{2pt}{0ex}}\mathrm{g}$

## Molarity and the concentration of solutions

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

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