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### 5.Molarity.

posted Mar 31, 2016, 1:55 PM by Upali Salpadoru   [ updated Jun 16, 2017, 4:05 PM ]
Fig,1  Making solutions.

Solid -Liquid      solutions are  homogeneous  mixtures

Here are a few terms used to describe solutions.

(i)  Dilute solution.: One with an excess of solvent.(Liquid)

(ii) Concentrated solution:-  One with an excess of solute dissolved. (Solid)

(iii) Saturated solution:  A solution that may not dissolve any more of the solute at the same conditions such as temperature and pressure.

(iv) Molar solution:- A solution where the concentration is known as the number of moles in a known volume  of the solution. For example one molar solution should have one mole of solute in a litre of solution.

The strength or concentration is given by the amount of solute dissolved in one litre of the solution, which is usually water.  This may be given in several ways. The two important methods are here.

I.    Concentration by mass per volume.

 Concentration= Mass / Volume                     Usual unit ..  .  (g /L) Cm  =  m/V

II.    Concentration by moles per volume.

 Concentration Molarity =  No. of  Moles  / Volume                     Usual unit ..  (mol / L) M  =  n/V

n  =  number of moles in the solution.

Example 1.

A solution of volume 0.250 was made by dissolving 5.3 g of sodium carbonate.

i.                  Find the strength of the solution in grams per litre.

ii.              Find the strength of the solution in mol/ L.

i.                  :-     0.250 L of solution has  …..5.3 grams

1L ……………………=5.3 / 0.25    = 21.2g/L.

ii.               Molar mass of Na2CO3 = 46+ 12+48 =  106 g/mol

106 g is………………….= 1 mol / L

21.2 g …………………  = 21.2/ 106

=0.2 mol / L

Example 2

A student took 0.25 L of a sodium hydroxide solution of 1-M strength.   Find the strength in grams per litre.

1-M 1 L will have 40g

1-M 0.25 will have   40x 0.25 = 10 g.

Diluting a  concentrated solution.

There are concentrated  solutions made as stock samples.  Very often we have to take a small volume from that and dilute it to a suitable standard. Here is a problem like that.

WARNING

We must never add water to acid.

In diluting acids we always add acid to water

Example 3

Find the volume of water that has to be added to 0.025 L of a  5-M oxalic acid solution to dilute it to a  1.0- M solution. Oxalic acid = H2C2O4.

In this case the easiest path is to use the formula.

 M1  V1   =   M2 V2

M2V2   =   M1V1

1 x V2=  5 x .025

Y2 = 0.125

As the new volume has to be 0.125 L and already there is 0.025 L.

the volume of water to be added =  0.1 L

 Ag Al Ar Au B Ba Be Br C Ca Cd Cl Cu Cr Fe H He Hg 108 27.0 40.0 197 10.8 137 9.0 79.9 12.0 40.1 112 35.5 63.6 52.0 55.9 1.01 4.0 201

 I K Li Mg Mn N Na O P Pb S Sn Si Zn 127 39.1 6.9 24.3 54.0 14.0 23.0 16.0 31.0 207 32.1 119 26.1 119

1.0  This question is only about Sodium hydroxide.  NaOH.

Find the following:

i.              Mass required to make 1L  of 1M solution.

ii.           Mass required to make  1L of 0.5 –M. solution.

iii.         Mass of solute present in 0.5 L  of a 1-M solution.

iv.         Mass of solute present in 0.5 L  of a 0.5-M solution.

v.            Volume of solution to dissolve 10g and make a 1-M solution.

vi.         Volume of solution to dissolve 10g and make a 0.5 M solution.

vii.       Mass of solute in 25 ml  of  a 2-M solution

(4x5=20 points)

i. 40g.     ii.  20g.   iii. 20 g.   iv. 10g   v.. 0.250 L  vi.  0.5 L

vii.   1000 ml has   80g.

25ml  will have     80x25 /1000   =  2g.

2.0 This question is about sodium carbonate. Na2CO3

Find the following

i.              The mass required to make 1 L of 1-M solution.

ii.           The mass required to make 250ml of 1M solution.

iii.         The molarity of a solution  made by dissolving 318g in 1L.

iv.         The mass of  Na2CO3 .10.H2O  for 1L of  1M solution.

(5x4 = 20 points)

i.     106g.

ii.    26. 5 g

iii.   1.L has    318g

1 L has    ……318/106 moles.  = 3 moles.

Molarity of solution is = 3.

iv.   Molar mass of  Na2CO3 .10.H2O  .

That is   106 +        18 ==   124 g/mol.

Mass  required is 124g.

3.0          This question is about potassium hydroxide.  KOH.

The mass of  residue after evaporating 50 ml. of a solution      was 5.6grams.

Find the following-

i.              The strength of the solution as grams per litre.

ii.           Molarity of the solution.

iii.         Number of molecules in the 1L of solution.

iv.         Number of K+ ions in 10 ml of the solution.

(5x4=20 points)

i.50 ml of solution had … 5.6g

ii.1000 ml (1 L)   …………5.6x1000 / 50.    = 56x2  =     112 g.

iii.Converting 112g to moles   112/molar mass  =   112/56   = 2 mol.

iv.The  solution is 2 M.molar.

4.0     This question is about sulphuric acid.  H2SO4.

Find the following.

i.               Number of moles in 0.025 L  taken from a 6-M solution.

ii.            How much water can be added to it to dilute it to half it’s strength?

iii.          Volume of acid that has to be added  1L of water to make a  1M solution?

iv.         How many H+ ions would be present in 1L of a 1-M solution?

(5x4= 20)

i.   One L solution has  6 moles.

0.025 L………….-6x0.025 mol.

=        0.15 moles.

ii. 0.  (water should never be added to acids)

iii.M2V2 =M1V1

3 x V2 = 6x 1

V2  =  6/3 =  2  L

As the new volume has to be 2.L,

amount to be added is 1.L.

iv.One L  of 1M Acid will have  1 mol of molecules.

One mol. Of  acid has  =  AN  molecules.

Each molecule has 2 H+ ions.

Therefore   the no. of ions  =  2 AN

5.0     This question is about diluting solutions.

Find the following:-

i.              Volume of water that has to be added to a 50 ml of a 1-M solution of glucose  to dilute to a Deci molar solution. ( deci = 1/10 )

ii.           What will be the molarity of a mixture  when  25 ml of a 2-M acid solution is mixed with 25 ml of a 1-M acid solution?

iii.         50 ml of a 2M- KOH solution was mixed with 25ml of a 3-M .KOH solution. What is the molarity of the mixture.

iv.         50 ml of Normal  saline NaCl aq is 0.154M..  How much water should be added to 100ml of 1M solution to bring it into Normal saline.

(sx4=20)

i.              M2V2 =M1V1

0.1x V2 =  1 x 50

V2 …..=  50 /0.1

=  500 ml.   As the new volume is 500 ml  , amount to be added is 450 ml.

ii.           2M solution will yield   1000ml……2mol

,                        25ml                    2x25 /1000   =  0.05mol  =

1M  solution  will have   1000ml…..1mol

25…..25/1000    =  0.025  mol

50 ml will have ………………………………….0.075

1000ml…………………….0.075x1000/ 50 =  1.5 mol

The  solution is =  1.5-M.

iii.         50ml of 2M-KOH has …..=  2  x50/1000 =0.1 mol.

25ml of 3M ……..= 3x25/1000  =  0.075 mol

Total 75 ml   will have     =  0.1 + 0.075  =   0.175mol in 75 ml.

In 1000  ml………= 0.175x1000/75  = 2,33 mol.

Strength of solution is =  2.33 M.

Iv     M2V2 = M1V1

0.154x V2 =  1x 50

V2=  50/0.154  =  324.68 ml

Volume to be added ..=  324.68- 50  =274.68 ml.