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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 5
00 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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