You are here: Home Amount of substance

## Molar Volumes of Gases

Definition 1: Molar volume of gases

One mole of gas occupies 22,4 dm3 at standard temperature and pressure.

### Note:

Standard temperature and pressure (S.T.P.) is defined as a temperature of 273,15 K and a pressure of 0,986 atm.

For example, 2 mol of $H 2$ gas will occupy a volume of 44,8 dm3 at standard temperature and pressure (S.T.P.). and 67,2 dm3 of ammonia gas ($NH 3$) contains 3 mol of ammonia.

## 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.

Solutions and moles

$C=n V$(1)

For this equation, the units for volume are dm3 (which is equal to litres). Therefore, the unit of concentration is mol·dm−3.

Definition 2: Concentration

Concentration is a measure of the amount of solute that is dissolved in a given volume of liquid. It is measured in mol·dm−3.

### Example 1: Concentration calculations 1

#### Question

If 3,5 g of sodium hydroxide ($NaOH$) is dissolved in 2,5 dm3 of water, what is the concentration of the solution in mol·dm−3?

(2)
##### Calculate the concentration
(3)

The concentration of the solution is 0,035 mol·dm−3.

### Example 2: Concentration calculations 2

#### Question

You have a 1 dm3 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 mol·dm−3?

##### Calculate the number of moles

$C=n V$ therefore:

(4)
##### Find the mass

The mass of $KMnO 4$ that is needed is 31,6 g.

### Example 3: Concentration calculations 3

#### Question

How much sodium chloride (in g) will one need to prepare 500 cm3 of solution with a concentration of 0,01 mol·dm−3?

##### Find the mass

The mass of sodium chloride needed is 0,29 g

### Exercise 1: Concentration of solutions

5,95 g of potassium bromide was dissolved in 400 dm3 of water. Calculate its concentration.

We use $C=\frac{n}{V}$ to calculate the concentration.

First we need to find n:

The molecular formula for potassium bromide is KBr. The molar mass of potassium bromide is:

$39,1+79,9=119\text{g}\cdot {\text{mol}}^{-1}$

The number of moles is:

 $n=\frac{m}{M}$ $n=\frac{5,95}{119}$ $n=0,05\text{mols}$

We need to also convert the volume to the correct units:

$\frac{400c{m}^{3}}{1000}=0,4d{m}^{3}$

Finally we work out the concentration:

 $C=\frac{n}{V}$ $C=\frac{0,05}{0,4}$ $C=0,125\text{mol}\cdot {\text{dm}}^{-3}$

100 g of sodium chloride ($NaCl$) is dissolved in 450 cm3 of water.

1. How many moles of $NaCl$ are present in solution?

2. What is the volume of water (in dm3)?

3. Calculate the concentration of the solution.

a) The molar mass of sodium chloride is:



The number of moles of sodium chloride is:

   

b) To convert fro to   we divide by 1 000:



c) The concentration is:

  $C=\frac{1,7}{0,45}$ $C=3.801919969\text{mol}\cdot {\text{dm}}^{-3}$

What is the molarity of the solution formed by dissolving 80 g of sodium hydroxide ($NaOH$) in 500 cm3 of water?

The molar mass of sodium hydroxide is: .

The number of moles of sodium hydroxide is:

 $n=\frac{m}{M}$ $n=\frac{80}{39,99}$

The concentration is:

 $C=\frac{n}{V}$ $C=\frac{2}{0,5}$

What mass (g) of hydrogen chloride ($HCl$) is needed to make up 1000 cm3 of a solution of concentration 1 mol·dm−3?

$\frac{1000c{m}^{3}}{1000}=1d{m}^{3}$

The number of moles of hydrogen chloride is:

 $n=CV$ $n=\left(1\right)\left(1\right)$ $n=1mol$

The molar mass of hydrogen chloride is:

$1,01+35,45=36,46g\cdot mo{l}^{-1}$

The mass of hydrogen chloride needed is:

 $m=nM$ $m=\left(1\right)\left(36,46\right)$ $m=36,46g$

How many moles of $H 2 SO 4$ are there in 250 cm3 of a 0,8 mol·dm−3 sulphuric acid solution? What mass of acid is in this solution?

$\frac{250c{m}^{3}}{1000}=0,25d{m}^{3}$

The number of moles is:

 $n=CV$ $n=\left(0,8\right)\left(0,25\right)$ $n=0,2mols$

The molar mass of ${H}_{2}S{O}_{4}$  is:

$2\left(1,01\right)+4\left(15,99\right)+32,07=98,05g\cdot mo{l}^{-1}$

The mass of ${H}_{2}S{O}_{4}$  is:

 $m=nM$ $m=\left(0,2\right)\left(98,05\right)$ $m=19,61g$