Chemistry Solutions, Types of Solutions and Expression of Concentration of Solutions :

Topics Covered :

● Introduction
● Types of Solutions
● Mass percentage (w/w)
● Volume percentage (v/v)
● Mass by volume percentage (w/v)
● Parts per million
● Mole Fraction
● Molarity (M)
● Molality (m)

Introduction :

`=>` In normal life we rarely come across pure substances.

`=>` Most of these are mixtures containing two or more pure substances.

`=>` The properties of brass (mixture of copper and zinc) are quite different from those of German silver (mixture of copper, zinc and nickel) or bronze (mixture of copper and tin).

`=>` 1 part per million (ppm) of fluoride ions in water prevents tooth decay, while 1.5 ppm causes the tooth to become mottled and high concentrations of fluoride ions can be poisonous.

`=>` Sodium fluoride is used in rat poison.

Types of Solutions :

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`

`color{green} ("Homogeneous Solutions")` : These are mixtures of two or more than two components. These have uniform composition and properties throughout the mixture.

`color{green}("Solvent")` : The component that is present in the largest quantity is known as solvent. Solvent determines the physical state of solution.

`color{green}("Solute")` : One or more components present in the solution other than solvent are called solutes.

`color{green}("Dilute solution")` : A solution in which relatively a small amount of solute is dissolved in large amount of solvent is called a dilute solution.

`color{green}("Concentrated solution")` : A solution in which relatively a large amount of the solute is present is called a concentrated solution.

`color{green}("Note") :` We shall consider only binary solutions (two component solutions) in this unit.

Expressing Concentration of Solutions :

`=>` Composition of a solution is described by expressing its concentration.

`=>` It is done either qualitatively or quantitatively.

`=>` For example, qualitatively we can say that the solution is dilute (i.e., relatively very small quantity of solute) or it is concentrated (i.e., relatively very large quantity of solute).

There are several ways by which we can describe the concentration of the solution quantitatively

Mass percentage (w/w) :

The mass percentage of a component of a solution is defined as :

`color{green}("Mass % of a component") = text(Mass of the component in the solution)/text(Total mass of the solution) xx 100`

`color{red}("Example") :` If a solution is described by `10%` glucose in water by mass, it means that `10` g of glucose is dissolved in `90` g of water resulting in a `100` g solution.

`=>` It is commonly used in industrial chemical applications. For example, commercial bleaching solution contains `3.62` mass percentage of sodium hypochlorite in water.

Volume percentage (v/v) :

The volume percentage is defined as :

`color{green}("Volume % of a component") = text(Volume of the component)/ text(Total volume of solution) xx 100`

For example, 10% ethanol solution in water means that 10 mL of ethanol is dissolved in 90 mL water such that the total volume of the solution is 100 mL.

`=>` Solutions containing liquids are expressed in this unit.

`=>` A 35% (v/v) solution of ethylene glycol, an antifreeze, is used in cars for cooling the engine. At this concentration the antifreeze lowers the freezing point of water to `255.4K` (`-17.6°C`).

Mass by volume percentage (w/v) :

It is commonly used in medicine and pharmacy. It is the mass of solute dissolved in 100 mL of the solution.

Parts per million :

When a solute is present in trace `color{green}("i.e. very small quantities")` quantities, then it is expressed in parts per million (ppm) and is defined as :

`color{green}("Parts per million ")` : `text(Number of parts of the component)/text(Total number of parts of all components of the solution) xx 10^6`

`=>` Concentration in parts per million can be expressed as mass to mass, volume to volume and mass to volume.

`=>` 1 litre of sea water (which weighs 1030 g) contains about `6 × 10^-3` g of dissolved oxygen (`O_2`). Such a small concentration is expressed as `5.8` g per `10^6` g (5.8 ppm) of sea water.

`=> color{red}("The concentration of pollutants in water or atmosphere is often expressed in terms of ")` `color(red)( μg mL^(-1) text{or ppm.}") `

Mole Fraction :

`=>` Commonly used symbol for mole fraction is`color{red}(x)`and subscript used on the right hand side of `color{red}(x)`
denotes the component.

It is defined as :

`color{green}("Mole fraction of a component:")` = `text(Number of moles of the component)/text(Total number of moles of all the components)`

For example, in a binary mixture, if the number of moles of `A` and `B` are `n_A` and `n_B` respectively, the mole fraction of `A` will be

`color{red}(x_A = n_A/(n_A+ n_B))`

For a solution containing `i` number of components, we have :

`color{red}(x_i = n_i/(n_1 + n_2 + n_3 + ...........+ n_i) = n_i/(sum n_i))`

In a given solution sum of all the mole fractions is unity, i.e.

`color{red}(x_1 + x_2 + .................. + x_i = 1)`

`=>` Mole fraction unit is useful in relating some physical properties of solutions, for example vapour pressure with the concentration of the solution and quite useful in describing the calculations involving gas mixtures.
Q 2955178064

Calculate the mole fraction of ethylene glycol (`C_2H_6O_2`) in a solution containing `20%` of `C_2H_6O_2` by mass.
NCERT
Solution:

Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g of water.

Molar mass of `C_2H_6O_2` = `12 × 2 + 1 × 6 + 16 × 2 = 62 g mol^-1`.

Moles of `C_2H_6O_2` = `(20 g)/(62 g mol^(-1))`

`= 0.322 mol`

Moles of water = `(80 g)/(18 g mol^(-1))`

`= 4.444 mol`

`x_text(glycol) = (text{moles of} C_2 H_6 O_2)/(text{moles of} C_2 H_6O_4 + text {moles of} H_2O)`

= `(0.322 mol)/(0.322mol + 4.444mol) = 0.068`

Similarly, `x_text(water) = (4.444 mol)/(0.322 mol + 4.444 mol) = 0.932`

Mole fraction of water can also be calculated as : `1 – 0.068 = 0.932`

Molarity (M) :

It is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

`color{green}("Molarity")` = `text(Moles of solute)/ text(Volume of solution in litre)`

For example, `0.25 mol L^-1` (or `0.25 M`) solution of `NaOH` means that `0.25` mol of `NaOH` has been dissolved in one litre (or one cubic decimetre).
Q 2985578467

Calculate the molarity of a solution containing `5 g` of `NaOH` in `450 mL` solution.
NCERT
Solution:

Moles of `NaOH = (5g)/(40 g mol^-1) = 0.125 mol`

Volume of the solution in litres = `450 mL // 1000 mL L^-1` Using equation `(2.8)`,

Molarity = `(0.125 mol xx 1000 mL L^–1)/(450 mL) = 0.278 M`

`= 0.278 mol L^–1`

`= 0.278 mol dm^–3`

Molality (m) :

It is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as

`color{green}("Molality (m)")` = `text(Moles of solute)/text(Mass of solvent in kg)`

For example, `1.00 mol kg^-1` (or `1.00` m) solution of `KCl` means that 1 mol (74.5 g) of `KCl` is dissolved in 1 kg of water.

`=>` Mass %, ppm, mole fraction and molality are independent of temperature.

`=>` Molarity is dependent of temperature because volume depends on temperature and the mass does not.
Q 2965778665

Calculate molality of `2.5 g` of ethanoic acid `(CH_3COOH)` in `75 g` of benzene
NCERT
Solution:

Molar mass of `C_2H_4O_2: 12 xx 2 + 1 xx 4 + 16 xx 2 = 60 g mol^–1`

Moles of `C_2H_4O_2 = (2.5 g)/(60 g mol^−1) = 0.0417 mol`

Mass of benzene in kg `= 75 g//1000 g kg^–1 = 75 × 10^–3 kg`

Molality of `C_2H_4O_2 = (text{Moles of} C_2 H_4 O_2)/(text{kg of benzene}) = (0.0417 mol xx 1000 g kg^-1)/(75 g)`

`= 0.556 mol kg^–1`

 
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