Methods of Expressing the Strength of Solution

Introduction

A solution is a homogeneous mixture of two (or more) substances, the composition of which may vary between certain limits. This concept is valid if we take the example of alcohol and water i.e. both the substances are in same phase.

But what about sugar syrup. Which contains 60% sugar and 40% water ? Will the sugar be the solvent ? Answers is No, Still the sugar is solute. Then what is the right definition for this case. Well, for hetrogenous substance,

” Solvent is that component of the binary mixture which is in some physical state as the solution will be ”

A solution consisting of two components is called binary solution. The component which is present in large quantity is called solvent and the component which is small in quantity is called solute. If both components are in same physical state.

Methods of Expressing the Strength of Solution

Mass Percentage: It may be defined as number of parts by mass of solute per hundred parts by mass of solution.

$\large Mass \; Percentage = \frac{Wt. \; of \; Solute}{Wt.\; of\; Solution }\times 100$

Volume Percentage: It may be defined as number of parts by volume of solute per hundred parts by volume of solution.

$\large Volume \; Percentage = \frac{Volume \; of \; Solute}{Volume \; of\; Solution }\times 100$

Molality: Molality of a solution is defined as the number of moles of solute dissolved in 1 Kg of the solvent.

$\large Molality \; m = \frac{moles \; of \; Solute}{Wt.\; of\; Solvent(in\; kg) }$

Molarity: The molarity of a solution gives the number of gram molecules of the solute present in one litre of the solution. Thus, if one gram molecule of a solute is present in 1 litre of the solution, the concentration of the solution is said to be one molar.

$\large Molarity \; M = \frac{no. \; of \; moles \; of \; Solute}{Volume\; of\; Solution(in\; litre) }$

Normality: The normality of a solution gives the number of gram equivalents of the solute present in one litre of the solution. Thus, if one gram equivalent of a solute is present in one litre of the solution, the concentration of the solution is said to be one normal.

$\large Normality \; N = \frac{no. \; of \; gram \; equivalent \; of \; Solute}{Volume \; of\; Solution(in\; litre) }$

Mole Fraction: The mole fraction of any component in a solution is the ratio of the number of moles of that component to the total number of moles of all components present in the solution.

$\large X_{solute} = \frac{n}{n+N} $

$\large X_{solvent} = \frac{N}{n+N} $

n = moles of solute and N = moles of solvent.

Note: Sum of the mole fractions of all the components in the solution is equal to unity.

Illustration : Concentrated hydrochloric acid contains 37% (by mass) HCl. The density of its solution is 1.18 g/ml. Calculate the molarity and molality of the solution.

Solution: Mass of the solution = 100 g

Since density, ρ = mass / volume

∴ Volume of the solution = mass / density = 100 g/ 1.18 g ml−1

= 85 ml = 85 cm3 = 0.085 dm3

n(HCl) = 37 g/36.5 g mol−1 = 1.01 mol

∴ Molarity of HCl = 1.01 mol/0.085 dm3 = 12.0 mol dm−3 = 12.0 M

Mass of solvent = (100 − 37) = 63 g = 0.063 kg

∴ Molality of HCl = 1.01 mol/0.063 kg = 16.03 mol kg−1 = 16.03 m

Also Read :

→ Vapour Pressure of Solution
→ Ideal and Non – Ideal Solutions
→ Colligative Properties
Measurement of Relating Lowering of Vapour Pressure
→ Boiling Point Elevation by a Non-Volatile Solute
→ Depression of Freezing Point by a Non-Volatile Solute
→ Osmosis and Osmotic Pressure
→ Abnormal Molecular Weight & Van’t Hoff Factor
→ Dissociation & Degree of Dissociation
→ Surface Tension
→ Relation b/w surface energy and surface tension
→ Angle of contact
→ Capillarity
→Viscosity

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