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Moles

Concentrations of Solutions - Mass and Moles.

Concentrations can be expressed as mol/dm³ or grams per dm³.
You can convert one into the other.

Example 1.

120 g of ammonium nitrate were dissolved in 1 litre of water.
What is the concentration of ammonium nitrate in mol/dm³?

Method.
1)  Work out the concentration in grams per dm³.
1 litre = 1 dm³ = 1000 cm³.

120 g per litre = 120 g per dm³.

2) Convert grams per dm³ into mol/dm³.
To convert mass into moles use  moles = mass ÷ RFM

RFM of ammonium nitrate (NH₄NO₃) is
(1 x 14) + (4 x 1) + (1 x 14) + (3 x 16)
= 80

moles = 120 ÷ 80
  = 1·5

So, 120 g of ammonium nitrate
dissolved in 1 litre of water = 1·5 mol/dm³.

Example 2.

10·6 g of sodium carbonate were dissolved in 200 cm³ of water.
What is the concentration of sodium carbonate in mol/dm³ ?

Method.
1)  Work out the concentration in grams per dm³.
1 litre = 1 dm³ = 1000 cm³.

10·6 g of sodium carbonate in 200 cm³ is equivalent to
10·6 x (1000 ÷ 200) g of sodium carbonate in 1000 cm³
= 53 g per dm³.

2)  Convert grams per dm³ into mol/dm³.
To convert mass into moles use  moles = mass ÷ RFM

RFM of sodium carbonate (Na₂CO₃) is
(2 x 23) + (1 x 12) + (3 x 16)
= 106

moles = 53 ÷ 106
  = 0·5

So, 10·6 g of sodium carbonate
in 200 cm³ of water = 0·5 mol/dm³.

Example 3.

How many grams of sodium hydroxide are present
in 500 cm³ of 0·8 mol/dm³ sodium hydroxide solution.

Method.
1) Convert mol/dm³ into grams per dm³.
To convert moles into mass use   mass = moles x RFM.

RFM of sodium hydroxide (NaOH) is
(1 x 23) + (1 x 16) + (1 x 1)
= 40

mass = 0·8 x 40
  = 32

0·8 mol/dm³  = 32 grams per dm³.

2)  If there are 32 grams of sodium hydroxide in 1000 cm³,
there are 32 x (500 ÷ 1000) g of sodium hydroxide in 500 cm³
= 16 grams of sodium hydroxide.

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