Solution! Even though you might not know its proper definition, but you use it frequently in your daily life. Chocolate syrup, sugar syrup, deodorants, perfumes, paints, nail paints, milk, tea, coffee, air, blood, and sweat are some examples of solutions. Did you notice any similarity among them? Yes, these are all mixed so well that we cannot separate their ingredients. It means that solutions are homogeneous mixture of two or more substances. There is one more similarity among above examples, they have one main substance and other substances are dissolved in it, for example, in sugar syrup, sugar is dissolved in water. The substance which is present in larger quantity is called the solvent and other substances which are present in lesser quantities are known as solutes. There are three states of any substance, solid, liquid and gas. Solvent and solute both can be solid, liquid or gas, there are no restrictions. It means we can have 9 combinations or solutions. Let’s have a look.
Solution | Solvent | Solute | Example |
Solid solution | Gold(solid) | Copper(solid) | This solution is used for jewelry making |
Silver(s) | Mercury(l) | Used in dental filling | |
Palladium(s) | Hydrogen(g) | Use as catalyst | |
Liquid solution | Water(l) | Sugar(s) | Sugar syrup |
Water(l) | Ethanol(l) | Alcoholic beverages | |
Water(l) | CO2(g) | Carbonated drinks | |
Gaseous solution | Air(g) | Particles(s) | Smoke |
Air(g) | Water(l) | Fog you have seen in the morning | |
Nitrogen(g) | O2(g) | Air is the solution of all gases |
Every solution has its unique properties such as taste, texture, viscosity, smell and so on. For example, beer, whisky, wine, and vodka are all alcoholic beverages, but every one of them is different from others. Each one has different alcohol content. Sugar syrups used for different purposes have different sugar content. It means that the ratio of solute and solvent plays an important role and there are a number of ways to represent this ratio or the concentration of solution. "How much solute is dissolved in how much solvent?", let’s see how many different ways and units are there to answer this question:
Weight percentage (w/w)
Weight percentage of a solute = Weight of solute in the solution × 100
Total weight of solution
If it is said that it’s a 10% glucose solution. That means 10g glucose is dissolved in 90g of water.
Mass percentage
Mass percentage of a solute = Mass of solute in the solution × 100
Total mass of solution
let’s try to understand it by taking an example of water. Find out the mass percentage of Oxygen in the water molecule.
mass of oxygen is 16 and molecular mass of water is 18 so,
Mass percentage of a Oxygen = 16 × 100
18
= 88.88%
Volume percentage (v/v)
Volume percentage of a solute = Volume of solute in the solution × 100
Total volume of solution
In laboratory we often use this unit of concentration. For example, 3% (v/v) hydrogen peroxide H2O2solution means 3ml H2O2 is dissolved in water such that the total volume of solution is 100ml.
Weight by Volume percentage (w/v)
w/v % of a solute = weight of solute in the solution × 100
Total volume of solution
This unit commonly used in pharmacy. Here we measure the weight of solute in grams dissolved in 100ml of solution. For example: 5% (w/v) saline solution is means 5g NaCl is dissolved in water to make 100ml of solution.
Parts per million (ppm)
You must have heard this term when talking about pollutants in air or water. When a solute is present in very little quantity then its concentration is expressed in ppm.
Parts per million solute = Number of parts of the solute in the solution × 106
Total number of parts of all components of the solution
For example, 6×10-3g of oxygen O2 is dissolved in 1liter of sea water ( which weighs 1030g), let's express its concentration in ppm.
Parts per million O2 = 6×10-3g × 106
1030
= 5.8ppm
Mole fraction(x)
Mole fraction of a component = Number of moles of the component in the solution
Total number of moles of all components of the solution
you can calculate the number of moles by dividing the given weight by its molecular weight.
number of moles = given weight in g / molecular weight
let’s take an example: suppose there is a solution of A, B and C and the number of moles of A, B and C are nA, nB and nCrespectively.
Mole fraction of component A (xA) = nA
( nA + nB + nC)
Similarly you can calculate the mole fraction of B and C. The sum of the mole fractions of all the components of a solution is always unity. let’s check it:
xA + xB + xC = ?
On putting the mole fractions in the above equation
nA + nB + nC
( nA + nB + nC) ( nA + nB + nC) ( nA + nB + nC)
nA + nB + nC
( nA + nB + nC)
= 1
So, xA + xB + xC = 1
Mole itself is an important measuring unit and mole fraction is useful when we talk about gaseous mixture or solutions of different components. You will find its applications when we discuss the properties of solutions.
Molarity (M)
In laboratory you must have seen the sign ‘M’ on reagent bottles. It is the number of moles of solute dissolved in one liter of solution.
Molarity M = Number of moles of solute
Volume of solution liter
or,
Molarity M = Number of moles of solute ×1000
Volume of solution ml
For example 2M HCl solution means 2moles of HCl is dissolved in 1liter. Let’s calculate the molarity of a solution containing 5g NaOH in 500ml solution.
Molecular weight of NaOH = 40
40g NaOH contains 1 mole
so 5g NaOH will contain = 5g/ 40g = 1/8 mole
and given volume of the solution is 500ml
M = 1 ×1000
8 × 500
M=1×2/8 =1/4
M = 0.25M
Molarity of a solution depends on temperature because volume of the solution may vary with temperature.
Normality (N)
It is the number of equivalent weight present in 1 liter of solution. Equivalent weight is the amount of substance which gives 1mol of H+ or 1mol of electrons. Let's try to understand it in a simpler way, 1 mol of HCl can give 1mol H+ so its equivalent weight will be same as its molecular weight. 1 mol H2SO4 can give 2mol H+ so its equivalent weight will be half of its molecular weight.
Normality (N) = Number of equivalents of solute
Volume of solvent in L
Number of equivalents = given weight / equivalent weight
For example 1 N solution of HCl means that 1equivalent of HCl is dissolved in 1L solution.
Q. calculates the normality of solution containing 7.88g of HNO3 per liter solution.
Equivalent weight of HNO3 = molecular weight of HNO3 = 63.02
number of equivalents in 7.88g = 7.88 / 63.02 = 0.125 eq
Normality (N) = Number of equivalents of solute
Volume of solvent in L
N = 0.125 / 1L
Answer = 0.125N
We will practice more problems on normality when we study redox reactions.
Molality (m)
It is defined as the number of moles of solute present in 1kg of solvent.
Molality (m) = Number of moles of solute
Mass of solvent in kg
for example 2m KI solution means that 2 moles of KI dissolved in 1kg of water. let’s calculate the molality of a solution containing 120g acetic acid in 100g water.
1 mole of acetic acid CH3COOH = 60g
number of moles in 120g = 120/60 = 2moles
Molality = 2 mole
0.1kg
answer: 20m
Molality is independent of temperature since mass doesn’t vary with temperature.
Let’s practice a few examples:
Q1. The density of a 2.03 M solution of acetic acid in water is 1.017 g/mL. Calculate the molality (m) of the solution.
To calculate the molality, first we will calculate the mass of acetic acid dissolved in 2.03M solution and then we will calculate the mass of solvent that is water.
Given solution contains 2.03 moles of acetic acid.
since molecular mass of acetic acid is 60g
then 2.03 moles = 2.03 × 60g = 121.80 g
In the given 2.03M solution 121.80 g acetic acid is dissolved in 1 L of solution. So we have to find out the weight of 1 L solution.
density of the solution is given 1.017 g/mL
Density = weight / volume
1.017 = wt / 1ml
So, wt = 1.017
1ml of solution weighs 1.017g
so 1000ml will weigh = 1.017× 1000 = 1017g
Now we know the weight of solution. But to calculate the molality we need to know the weight of solvent.
weight of solvent = weight of solution - weight of solute
=1017- 121.80
=895.2g = 0.8952kg
now we can calculate the molality of solution
molality = 2.03moles
0.8952kg
Answer = 2.26m
Q2. Calculate the molarity of a sulphuric acid H2SO4 solution of density 1.198g/cm3, containing 27% sulphuric acid by weight.
Hint: To calculate M, we need to find out the weight of H2SO4 present in given solution, then we can calculate the moles and then we calculate M.
it is given that,
1cm3solution weighs 1.198 g and it contains 27% H2SO4
= 27 × 1.198 = 0.32g H2SO4
100
molecular weight of H2SO4 = 98.1
number of moles present in 0.32g = 0.32/ 98.1 = 0.0033 moles
1cm3 = 1ml = 1× 10-3 L
M = number of moles of solute
volume in L
M = 0.0033
1 × 10-3
M = 3.3 mol/L = 3.3M
Q3. How many g of CaCl2 should be added to 300mL water to make 2.46m solution?
molecular weight of CaCl2 = 111
Molality (m) = Number of moles of solute
Mass of solvent in kg
density of water is 1 g/mL
so the weight of 300ml water will be = 300g = 0.3kg
Molality (m) = (weight/ molecular weight of solute)
Mass of solvent in kg
2.46m = (weight/ 111)
0.3kg
2.46m × 0.3kg × 111 = weight
Answer: 81.91g of CaCl2 should be added.
Q4. what is the mole fraction of the solute in 1m aqueous solution?
1m aqueous solution means 1mole of solute is dissolved in 1kg of water.
molecular weight of water = 18
number of moles of water in 1000 g = 1000/18 = 55.5 moles
Mole fraction (x) of a component = Number of moles of the component in the solution
Total number of moles of all components of the solution
x = 1
1+ 55.5
answer: 0.017
In the next post we will discuss more about solutions.
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