Concentration of Solution.
Concentration – The amount of solute present in a unit volume of a solution is called the concentration of the solution.
The concentration of a solution is expressed in the following terms.
Methods of measurement.
1. Mass percent (w/w)%
(w/w)% = mass of solute / mass of solution x 100
For example, 10% (w/w)% means a solution of glucose in water.
10 gm of glucose is dissolved in 90 gm of water.
2. Volume percent (v/v)%
(v/v)% = volume of solute / volume of solution x 100
For example, 10% (v/v)% means a solution of ethanol in water.
10 ml of ethanol is dissolved in 90 ml of water.
3. Mass-Volume Percentage (W/V) %
(W/V) % = Mass of solute / Volume of solution x 100
For example: Pharmacy (in medicines)
4. Parts per million (PPM). This term of concentration is used when the amount of solute is very small.
For example: 1. Dissolved oxygen in water
2. Amount of soluble salts in drinking water
PPM = Number of parts of solute / Number of total parts of solution x 106
5. Mole Fraction. The mole fraction of a component/component in a solution is equal to the ratio of the number of moles of that component to the total number of moles present in the solution.
Suppose the number of moles of two components A and B in a binary solution are nA and nB respectively, and the mole fractions of these components are xA and xB respectively. Therefore, the mole fraction of component A (XA) = nA/nA + nB
Similarly, the mole fraction of B (XB) = nB/nB + nA
Note: The sum of the mole fractions of all components in a solution is 1.
6. Molarity. The number of moles of solute present in 1 L of a solution is called molarity.
M = number of moles of solute / volume of solution (liters)
Question: A solution is prepared by dissolving 10g of NaOH in 250ml of water. Find the molarity.
⇒ Molarity = Number of moles of solute/Volume of solution (liters)
Molecular weight of NaOH:
28 + 16 + 1
40
Moles of solute (NaOH) = Weight/Molecular weight
= 10/40 = 1/4 mole
M = 1/4/250 x 1000
= 1/4 x 4
[M = 1 mole/L]
Question: Calculate the molarity of a solution containing 5g of NaOH dissolved in 450ml of solution.
⇒ Molarity = Number of moles of solute/Volume of solution (liters)
Molar weight of NaOH:
23 + 16 + 1
40
Moles of solute NaOH = Weight/Molecular weight
= 5/40 = 1/8 mol
M = 1/8/450 x 1000
= 1/36 x 10
= 5/18
[M = 0.27 mole/L]
7. Molality (m): The number of moles of solute present in 1 kg of solvent is called molality. It is denoted by m.
Or
The number of moles of solute dissolved in 1 kg of solvent is called molality.
Question: A solution is prepared by dissolving 10g of urea (NH2CONH2) in 500g of water. Find the molality of the solution.
⇒ Molality (M) = Number of moles of solute/Mass of solvent (kg)
(NH2CONH2) Molecular weight of urea:
14 + 2 + 12 + 16 + 14 + 2
60
Moles of NH2CONH2 = Mass/Molecular weight
= 10/60 = 1/6 mole
m = 1/6/500 x 1000
[m = 1/3 mole/kg]
Question: 0.05 mole of ethanoic acid is dissolved in 250 grams of benzene. Find the molality of the solution.
⇒ m = Number of moles of solute/Mass of solution (gm)
= 0.05/250 x 1000 = 1/5
m = 0.2 mole/kg
Question. Which of the following, molality or molarity, represents the concentration of a solution more accurately? Why?
Answer: Molality represents concentration more accurately because it is not affected by temperature.
Note: Mass percent, mole fraction, ppm, and molality (m) do not depend on temperature.
While molarity (M), mass-volume percent, and volume percent depend on temperature.
Question: Unit of molarity?
Answer: Mole/L
Question: A 35% (V/V) solution of ethylene glycol is used to cool vehicle engines. Find the volume of water in ml.
Answer: Ethylene glycol is 35% (V/V). Therefore, if the total volume is 100 ml, the volume of water will be 100 – 35 = 65 ml.
Therefore, the volume of water in this solution will be 65ml or 65%.
Question: Calculate the mole fraction of ethylene glycol (C2H6O2) if 20% of the mass of C2H6O2 is present in the solution.
⇒ Weight of C2H6O2 (ethylene glycol) = 20g
Assuming water is present as the solvent, then –
100 – 20 = 80
(Total solution = 100g)
Weight of ethylene glycol = weight/molecular weight
Molecular weight of C2H6O2:
24 + 6 + 32
62
Weight of C2H6O2 = 20/62
= 10/31 mole
= 0.322 mole
Moles of H2O = weight/molecular weight
Molecular weight of H2O
2 + 16 = 18
= 80/18
= 4.44 mole.
Mole fraction of ethylene glycol
XB = nB/nB+nA
=0.322/4.44+0.322
=0.068
Mole fraction of water
XA = nA/nA+nB
=4.44/4.44+0.322
=111/119
=0.932
