Moles and solutions

By the end of section you should be able to…

•Calculate the amount of substance in mol, using solution volume and concentration

• Describe a solution’s concentration using the terms concentration and dilute.

The concentration of a solution is the amount of solute, in mol, dissolve per 1 dm3 (1000 cm3) of solution. The amount, in moles, in a solution is:

n = c x V(in dm3)

or

n = c x V(in cm3) 1000

Worked examples1. What is the amount in moles of sodium

hydroxide in 50 cm3 of a solution of concentration 2 mol/ dm3

Answern = c x V (in cm3)

1000

n = 2 x 50 1000 = 0.01 mol

2. If 0.4 mol of sodium carbonate is dissolved to make 200cm3 of solution, what is its concentration?AnswerRe-arrange the equation n = c x V(dm3)

c = n V(dm3) c = 0.4 mol 0.20 dm3

c = 2 mol/dm3

3. What volume of a solution of hydrochloric acid of concentration 0.5 mol/dm3 contains 0.15 mol?

Give the answer in cm3

AnswerRe-arrange the equation n = c x V(dm3)

V(dm3) = n / c

= 0.15 / 0.5

= 0.3 dm3 To convert 0.3dm3 to cm3, multiply 1000

0.3 x 1000 = 300 cm3

Standard solution

A standard solution is a solution of known concentration.

Standard solutions are normally used in titration to determine unknown

information about another substance.

Mass concentration( g/dm3)

•Find the amount, in mol,required for 1 dm3 of the solution . .

•convert moles to grams, i.e from mass = n x M .

Find the mass concentration, in g/dm3 for the following solutions(a) 0.02 moles of HCl dissolved in 500cm3 of solution. (b) 0.25 moles of HNO3 in 2 dm3

of solution.

Answers(a) n = mole x 1 dm3 V

n = 0.02 x 1 dm3 0.5 dm3

n = 0.04 mol/dm3

From mass = n x M

M(HCl) = 1 + 35.5 = 36.5

mass conc. (g/dm3) = 0.04 x 36.5

= 1.46 g/dm3

(b) 0.25 moles of HNO3 in 2 dm3 n = mole x 1 dm3

V

n = 0.25 mol x 1 dm3

2

n = 0.125 mol/dm3

from mass conc. = n x M

M(HNO3) = 1 + 14+ (16x3) = 63

i.e. mass conc. = 0.125 x 63

= 7.88 g/dm3

NOW TRY QUESTIONS 1 -3

PAGE 17

Dilute solution

• A dilute solution is a solution with a small amount of solute per dm3

• Normal bench solutions of acids usually have concentrations of 1 mol/dm3 or

2 mol/dm3. These are dilute solutions.

Preparing dilute solution

The simple formula usually used is as follows:

M1V1 = M2V2

Where

M1 = original concentration

V1 = original volume

M2 = new concentration

V2 = new volume

How can 500 cm3 of a 0.4 mol/dm3 solution of sulphuric acid be prepared from a solution of

concentration 2 mol/dm3.

Answer 2 x V = 500 x 0.4volume = 500 x 0.4

2 = 100 cm3

therefore, 100 cm3 of the original solution be further diluted by adding 400 cm3 of water to

make 500 cm3 of 0.4 mol/dm3

What is the concentration in mol/dm3 of 375 cm3 of dilute sodium hydroxide solution prepared by diluting 75 cm3 of sodium hydroxide solution of concentration 0.6

mol/dm3 with water?

Answer From M1V1 = M2V2

75 x 0.6 = 375 x M2

M2 = 75 x 0.6

375concentration, M2 = 0.12 mol/dm3