What is the difference between mass percent and volume percent?

Mass percent (w/w): Expresses the concentration of a solute in a solution as the mass of solute per 100 grams of the solution. Volume percent (v/v): Expresses the concentration of a solute in a solution as the volume of solute per 100 milliliters of the solution.

How is molarity different from molality?

Molarity (M): Moles of solute per liter of solution. Molality (m): Moles of solute per kilogram of solvent.

3. What is normality and how is it related to molarity?

Normality (N): Number of gram equivalents of solute per liter of solution. Relation: Normality = Molarity × Valency factor

What is the significance of parts per million (ppm)?

ppm is used for expressing the concentration of very dilute solutions, such as those found in environmental samples or trace impurities.

How can you calculate the mass of solute given the molarity and volume of the solution?

Formula: Mass of solute (g) = (Molarity × Volume of solution (mL) × Molar mass) / 1000

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Unit of Concentration of Solutions

Units of Concentration of Solutions

The concentration of a solution is the amount of solute present in a given quantity of the solution or solvent. In a solution the relative proportion of the solute and solvent can be varied.

Depending upon the amount of solute present in a solution, it can be called a dilute, concentrated or a saturated solution.

1.0Concentration in Terms of Percentage

Mass By Mass Percentage (w/W)

The number of grams of solute that is dissolved in a given mass of solution in gram is called mass fraction of the solute. Thus,

$M a ss F r a c t i o n = \frac{W}{w + W}$

Where 'w' grams of solute is dissolved in W grams of solvent.

$M a ss p erce n t = \frac{M a ss o f so l u t e in g r am s \times 100}{M a ss o f so l u t i o n in g r am s} = \frac{w \times 100}{w + W}$

Volume By Volume Percentage (v/V)

This method is used for solutions of liquid in a liquid. The volume of liquid (solute) in ml dissolved in given volume of solution in ml is called volume fraction.

$V o l u m e f r a c t i o n = \frac{V o l u m e o f l i q u i d so l u t e in m l}{V o l u m e o f so l u t i o n in m l}$

$V o l u m e p erce n t = \frac{V o l u m e o f so l u t e \times 100}{V o l u m e o f so l u t i o n}$

Mass By Volume Percentage(w/v)

The mass of the solute in grams dissolved in 100 ml of the solution

$M a ss b y v o l u m e p erce n t a g e = \frac{M a ss o f so l u t e ( g )}{V o l u m e o f so l u t i o n ( m l )} \times 100$

2.0Molarity (M)

The number of moles of solute dissolved in one litre solution is called its molarity.

$M o l a r i t y = \frac{N u mb er o f m o l es o f so l u t e}{v o l u m e o f so l u t i o n in l i t res} = \frac{n}{V}$

$M o l a r i t y = \frac{N u mb er o f m o l es o f so l u t e \times 1000}{v o l u m e o f so l u t i o n in m l}$

$M o l a r i t y = \frac{M a ss o f so l u t e in g r am s \times 1000}{M o l a r ma ss \times V o l u m e o f so l u t i o n in m l}$

$M a ss o f so l u t e in g r am = \frac{M o l a r i t y \times V o l u m e o f so l u t i o n in m l \times m o l a r ma ss}{1000}$

3.0Normality (N)

The number of equivalents or gram equivalents of solute dissolved in one litre of the solution is known as normality (N) of the solution.

$N or ma l i t y (N) = \frac{N u mb er o f g r am e q u i v a l e n t s o f so l u t e}{V o l u m e o f so l u t i o n in l i t re}$

$= \frac{M a ss o f so l u t e in g r am}{Eq u i v a l e n t ma ss \times V o l u m e o f so l u t i o n ( l i t re )}$

$= \frac{St re n g t h o f so l u t i o n in g r am / l i t re}{Eq u i v a l e n t ma ss o f so l u t e}$

Equivalent mass of a substance is that mass which reacts with or displaces one gram of hydrogen, 8 grams of oxygen or 35.5 grams of chlorine.

4.0Relation Between Molarity and Normality

Molarity × Molar mass of solute = Normality × Equivalent mass of solute.

Normality =

$\frac{M o l a r i t y M o l a r M a ss o f S o l u t e}{Eq u i v a l e n t ma ss o f so l u t e}$ =

$\frac{M o l a r i t y \times M o l a r M a ss o f S o l u t e}{M o l a r ma ss o f so l u t e \times v a l e n cy f a c t or}$ =

$\frac{M o l a r i t y \times M o l a r M a ss o f S o l u t e}{\frac{M o l a r ma ss o f so l u t e}{v a l e n cy f a c t or}}$

Normality = Molarity × Valency factor

N = M × n;

N > M

5.0Molality (m)

The number of moles or gram molecules of solute dissolve in 1000 gram of the solvent is called molality of the solution.

Molality of a solution =

$\frac{N u mb er o f M o l es o f S o l u t e}{A m o u n t o f so l v e n t in k g .} = \frac{N u mb er o f M o l es o f S o l u t e \times 1000}{A m o u n t o f so l v e n t in g r am}$

6.0Formality (F)

It is the number of formula mass in grams present per litre of solution.

$F or ma l i t y = \frac{M a ss o f so l u t e in g r am}{F or m u l a ma ss o f so l u t e in g r am \times V o l u m e o f so l u t i o n in l i t re}$

7.0Concentration or Strength of a Solution (S)

The mass of solute in gram dissolved in one litre solution is known as its strength in gram per litre.

$S = \frac{M a ss o f S o l u t e in g r am s \times 1000}{V o l u m e o f so l u t i o n in m l}$

S = Molarity of solution × Molar mass of solute.

S = Normality of solution × Equivalent mass of solute.

8.0Parts Per Million (ppm)

This method is used for expressing the concentration of very dilute solutions such as hardness of water, air pollution etc.

$pp m o f s u b s t an ce = \frac{M a ss o f so l u t e}{M a ss o f so l u t i o n} \times 1 0^{6} = \frac{V o l u m e o f so l u t e}{V o l u m e o f so l u t i o n} \times 1 0^{6}$

9.0Mole Fraction

The ratio of the number of moles of one component to the total number of all the components present in the solution, is called the mole fraction of that component.

Mole fraction of solute X_A is given by

$X_{A} = \frac{n _{A}}{n _{A} + n _{B}}$

Mole fraction of solvent X_B is given by

$X_{B} = \frac{n _{B}}{n _{A} + n _{B}}$

where n_A is moles of solute A and n_B is moles of solvent B. (X_A + X_B = 1)

10.0Relation Between Molality and Molarity

$\frac{M o l a r i t y}{M o l a l i t y} = \frac{M o l es o f so l u t e}{V o l u m e o f so l u t i o n ( in L i t re )} \times \frac{M a ss o f so l v e n t ( kg )}{M o l es o f so l u t e}$

$= \frac{M a ss o f so l v e n t ( kg )}{V o l u m e o f so l u t i o n ( in l i t re )} = \frac{M a ss o f S o l v e n t ( in g r am s )}{V o l u m e o f so l u t i o n ( in m l )}$

Let the density of the solution be d g mL^–1

Mass of solution = V × d

Mass of solute = number of moles × Molar mass of solute

= n × m_A

Mass of solvent(W) = mass of solution – Mass of solute

= (V × d) – (n × m_A)

Thus,

$\frac{M o l a r i t y}{M o l a l i t y} = \frac{( v \times d ) - ( n \times m _{A} )}{V}$

$M o l a l i t y (m) = \frac{M o l a r i t y \times V}{( v \times d ) - ( n \times m _{A} )}$

$m = \frac{M o l a r i t y}{d - ( \frac{n}{v} \times m _{a} )}$

$m = \frac{M o l a r i t y}{d - ( m o l a r i t y \times m _{a} )}$

11.0Also Read

Method of Separation	Vapour Pressure	Physical Properties of Materials
Properties of Matter	The Importance of Learning Chemistry	Separation of Substance
States of Matter	Classification of Materials	Mixtures and Its Types

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