Practical expression of concentration

Practical Physical pharmacy

Lab. 1: expressions of concentration
-Dispersion system consists of solute (internal, dispersed) phase and solvent (external phase, dispersion media)
-examples of dispersion systems:
1-true solution: homogenous system
Solute(completely dissolved, cant scatter light)
2-colloidal dispersion : particle size of solute 1-500 nm
( ex: blood )
3-coarse dispersion:particle size of solute is more than 500nm
(ex :suspension, emulsion)
1) Molarity(M)=moles of solute / liters of solution M= mol / L

Moles of solute= Weight (gram) / Molecular weight (g/mol)

M.wt of NaCl (23+35)= 58 M.wt of H2SO4 = 2*1+32+16*4=98

2) molality (m) = moles of solute / kg of solvent

Changing temp. can change molarity but not molality so we use molality when there is a change in the temp. throughout the experiment.


3) mole fraction = number of moles ( part ) / number of moles (total)
EX: NaCl solution in water
Moles of NaCl =1 , moles of water = 4
So mole fraction of NaCl =1 / (1+4) = 1/5 and the mole fraction of water is 4 / (1+4) = 4 / 5
The sum of the mole fractions of all constituents must be 1

4) mole percent : mole fraction multiplied by 100

5) Normality (N)= no. of equivalent weights of solute / liters of solution

No. of equivalent weight of solute = weight (gram) / equivalent weight ,, equivalent weight=m.wt/n
For HCl , NaOH , NaCl , n= 1
CaCO3 , H2SO4 n=2
K3PO4 n = 3
-- Ca3(PO4 )2 ??
So when n=1 , m.wt = eq. wt

6) percent by weight ( % w/w) = grams of solute in 100 grams of solution

percent by volume ( %v/v) = milliliters(ml) of solute in 100 ml of solution
percent weight in volume (% w/v) = grams of solute in 100 ml of solution



Example / an aqueous solution of ferrous sulfate was prepared by adding 41.50 g of FeSO4 to enough water to make 1000 ml solution at 18 c .The density of the solution is 1.0375
(atomic weights : Fe=55.85 ,S= 32.06 ,O= 16 )
Calculate : a) molarity , b)molality , c)mole fraction of FeSO4 , mole fraction of H2O , mole percent of the two constituents d)percentage by weight of FeSO4 .

a) molarity

moles of FeSO4 = g of FeSO4 / m.wt m.wt=(55.85+32.06+16*4)=151.9 g/mol
= 41.50 /151.9 = 0.273
M = moles of FeSO4 / liters of solution
0.273 / 1 = 0.273 M
b) molality
grams of solution = volume * density
= 1000 * 1.0375 = 1037.5 g
Grams of solvent=grams of solution – grams of FeSO4
1037.5 – 41.5 = 996 g
Molality (m) = moles of FeSO4 / kg of solvent
= 0.273 / 0.996 = 0.2743 m
c) mole fraction and mole percent

moles of water = 996 / 18.02 = 55.27

mole fraction of FeSO4 = moles of FeSO4 / (moles of water + moles of FeSO4 )
0.273 /( 55.27 + 0.273) = 0.0049
Mole fraction of water = 55.27 / (55.27+0.273) = 0.9951
Mole percent of FeSO4 =0.0049 * 100 = 0.49 %
Mole percent of water = 0.9951 *100 = 99.51%


d) percentage by weight of FeSO4 = g of FeSO4 / g of solution * 100
41.50 / 1037.5 * 100 = 4.00 %
Practical :
1- prepare 0.1 M 100 ml of NaCl solution
2-prepare 0.1 N 100 ml of Na2CO3 solution
3-prepare 0.1 m of Na2SO4 (50 g)
4-prepare 5 mole percent of NaCl in water (100 ml )
5-prepare 40 % (v/v) of water/alcohol mixture (50 ml)