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Concentrations of Solutions - Mass and Moles.

Concentrations can be expressed as mol**/**dm^{3}
or grams per dm^{3}.

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**/**dm^{3}?

Method.

1) Work out the concentration in
grams per dm^{3}.

1 litre = 1 dm^{3} = 1000 cm^{3}.

120 g per litre = 120 g
per dm^{3}.

2)
Convert grams per dm^{3} into mol**/**dm^{3}.

To convert mass into moles
use moles = mass
÷ RFM

RFM of ammonium
nitrate (NH_{4}NO_{3}) 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**/**dm^{3}.

Example
2.

10·6 g of sodium carbonate were dissolved in 200 cm^{3} of water.

What is the concentration of sodium carbonate in mol**/**dm^{3} ?

Method.

1) Work out the concentration in
grams per dm^{3}.

1 litre = 1 dm^{3} = 1000 cm^{3}.

10·6 g of sodium
carbonate in 200
cm^{3} is equivalent
to

10·6 x (1000
÷ 200) g of sodium carbonate in 1000 cm^{3} = 53 g per dm^{3}.

2) Convert
grams per dm^{3} into mol**/**dm^{3}.

To convert mass into moles
use moles = mass
÷ RFM

RFM of sodium
carbonate (Na_{2}CO_{3}) 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
cm^{3} of water = 0·5 mol**/**dm^{3}.

Example 3.

How many grams of sodium
hydroxide are present

in 500 cm^{3} of
0·8
mol**/**dm^{3} sodium hydroxide
solution.

Method.

1) Convert mol**/**dm^{3} into grams per dm^{3}.

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**/**dm^{3} = 32 grams per dm^{3}.

2) If there are 32 grams of
sodium hydroxide in 1000 cm^{3},

there are 32 x (500
÷ 1000) g of sodium hydroxide in 500 cm^{3} =
16 grams
of sodium hydroxide.

Links Moles Revision Quizzes Revision Questions

gcsescience.com The Periodic Table Index Moles Quiz gcsescience.com

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