Interconversions of Concentration Terms - Mathematical derivations, Examples, Practice Problems & FAQs

If a person says 100 mL of your drinking water contains 0.00015 g of fluoride ion (F^-) or 0.00007 mol of fluoride ion present in 1 litre of your drinking water, you think these amounts can be neglected but if you informed that your 100 mL of drinking water contains 1.5 ppm of fluoride water then you think we have to worry about this concentration of fluoride ions. As per the study, fluoride ions in water at 1 part per million (ppm) prevent tooth decay, whereas 1.5 ppm produces tooth mottling and high amounts of fluoride ions can be deadly (for example, sodium fluoride is used in rat poison)

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Table of contents:

Concentration terms
Interconversions of concentration terms
Relation between molarity (M) & molality (m)
Relation between molality (m) & mole fraction ()
Relation between strength of solution and Molarity
Relation between strength of solution and Normality
Relation between normality of solution and molarity
Relation between $\frac{w}{v}$ % and molarity of solution
Practice problems
Frequently asked questions-FAQs

Concentration terms

There are different methods to represent the amount of solute present in a solution. E.g- Molarity, molality, normality, etc.

Temperature-dependant concentration terms

Molarity (M)
Normality (N)
$\frac{w}{v}$ %
$\frac{v}{v}$ %

Temperature independent concentration terms

Molality (m)
Mole fraction ()
$\frac{w}{w}$ %
Parts per million (ppm)

Molarity (M): It is defined as the moles of solute present in 1000 mL or 1 L of the solution is called molarity of the solution.

$M o l a r i t y (M) = \frac{m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n (L)} = \frac{m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n (m L)} \times 1000$

Molarity is temperature-dependent.
It varies inversely with temperature.
Mathematically: Molarity (M) $α \frac{1}{V o l u m e} α \frac{1}{T e m p r a t u r e}$

Case 1: Calculation of molarity, when solution having molarity (M1), diluted from V1 to V2

Initial molarity = M₁, Final molarity = M_F
Initial volume = V₁, diluted to volume = V₂

$\Rightarrow M_{F} = \frac{M_{1} V_{1}}{V_{2}}$

Case 2: Calculation of molarity when solution having molarity (M₁) & volume (V₁) mixed with a solution having the same solute with molarity (M₂) & volume (V₂)

$\Rightarrow M_{F} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}}$

Molality (m): molality is the number of moles of solute present in 1 kg or 1000 g of solvent.

$M o l a l i t y (m) = \frac{m o l e s o f s o l u t e}{w e i g h t o f s o l v e n t (k g)} = \frac{m o l e s o f s o l u t e}{w e i g h t o f s o l v e n t (g)} \times 1000$

Molality is temperature independent.

Normality(N): It is defined as the number of equivalents of solute present in 1000 mL or 1 L of the solution is called normality of solution.

$N o r m a l i t y (N) = \frac{n u m b e r o f e q u i v a l e n t o f s o l u t e}{v o l u m e o f s o l u t i o n (L)} = \frac{n u m b e r o f e q u i v a l e n t o f s o l u t e}{v o l u m e o f s o l u t i o n (m L)} \times 1000$

Parts per million (ppm): Parts per million is defined as the number of parts of the solute present in every 10⁶ parts of the solution.

E.g- 10 ppm solution means that 10 g of solute is present in every 10⁶ g of solution.

$P a r t s p e r m i l l i o n (p p m) = \frac{w e i g h t o f s o l u t e}{w e i g h t o f s o l u t i o n} \times 10^{6}$

= amount of solute in gram present in 100 mL of solution.

⁼
20 % solution means 20 g solute present in 100 g of solution.
=
20 % solution means 20 mL solute present in 100 mL of solution.
= amount of solute in gram present in 100 g of solution.
=
20 % solution means 20 g solute present in 100 mL of solution.
= amount of solute in mL present in 100 mL of solution.

Mole fraction ()

The ratio of the number of moles of the solute or solvent present in the solution and the total number of moles present in the solution is known as the mole fraction of substances concerned.

Let, the number of moles of solute in a solution = n1 & number of moles of solvent in a solution = n2

Mole fraction (x₁) of solute = $\frac{n_{1}}{n_{1} + n_{2}}$

Mole fraction (x₂) of solvent = $\frac{n_{1}}{n_{1} + n_{2}}$

For Binary solution; x₁ + x₂= 1

Strength of solution:

Strength of solution is the amount of solute in g present in 1000 mL of solution

$S t r e n g t h o f s o l u t i o n = \frac{w e i g h t o f s o l u t e (g)}{V o l u m e o f s o l u t i o n (m L)} 1000$

Interconversions of concentration terms

Relation between molarity (M) & molality (m)

m is the molality of solution

Weight of solute =
Weight of solvent =
Molar mass of solute =
Molar mass of solvent =
Molality (m) = —-------- (1)
Molarity (M) = —---------(2)
V = —------- (3)
—------- (4)
Weight of solution = w1 + w2 = w1+
Destiny of solution (ρ) = = =
Density of solution (ρ) = = =

Relation between molality (m) & mole fraction (x)

Let mole of solute present in 1000 g of solvent
m is molality of solution
weight of solute =
weight of solvent =
Molar mass of solute =
Molar mass of solute =
Mole of solute = =
Mole of solvent = =

Relation between strength of solution and Molarity

We know, Strength of solution is amount of solute in g present in 1000 mL of solution

Multiply equation (6) both numerator and denominator by Molar mass of solute

(Mole =
So,
Strength of solution (g L-1) = Molarity × molar mass

Relation between strength of solution and Normality

We know, Strength of solution is amount of solute in g present in 1000 mL of solution

Multiply equation (6) both numerator and denominator by equivalent weight of solute

(Number of equivalent = )

So,

Strength of solution ( g L -1 ) = normality × equivalent weight

Relation between normality of solution and molarity

Parts per million (ppm): it is defined as the number of equivalents of solute present in 1000 mL or 1 L of the solution is called normality of solution.

We know, (Number of equivalent = )

So,

Relation between $\frac{w}{v}$ % of solution and molarity

= amount of solute in gram present in 100 mL of solution.
20 % solution means 20 g solute present in 100 mL of solution.
=
Multiply equation (7) by (molar mass ×10) to both numerator and denominator
=
=
= Molarity ×

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Practice problems:

Q 1. Calculate the molality of the solution if the density of 1.25 M CH3COOH solution is 1.23 g mL^-1.

2.96 m
1.08 m
1.34 m
1.73 m

Answer: (B)

Solution: We know, Density of solution ( ρ ) =

Q 2. Find molality of an aqueous solution of glucose having mole fraction 0.1 and density of the solution is 1.1 g mol^-1.

6.17 M
3.85 m
5.34 m
4.73 m

Answer: (A)

Molar mass of solute (M1) = 180 g mL^-1

Molar mass of solute (M2) = 18 g mL^-1

Solution: We know, χ1 =

0.1 =

Molality (m) = 6.17 m

Q 3. Which will change with an increase in temperature?

Molality
Molarity
Both of these
None of these

Answer: (B)

Solution:

Molarity is temperature-dependent. it varies inversely with temperature.

mathematically: Molarity ( M )

Q 4. Find mole fraction of solute for 2 m NaOH solution.

0.887
0.143
0.012
0.035

Answer: (D)

Solution: molar mass of NaOH = 40 g mol^-1

2 m NaOH solution means, 2 moles of NaOH present in 1000 g of water

Moles of water = mol

Mole fraction of solute ( NaOH) =

Frequently Asked Questions-FAQs

Q1. Can the molality of any solution convert into molarity without knowing the density of the solution?

Answer: No, for conversion of molality into molarity or molarity into molality, density must be required. Molarity is volume-dependent and molality is only mass-dependent.

The density of solution (ρ) =

Q2. Can we consider molarity equal to molality for an infinitely dilute solution?

Answer: Only for numerical simplicity for a very dilute aqueous solution, we can consider molarity = molality. (only for aqueous solution)

We know, the density of water = 1 g cc^-1. So, for very dilute solutions weight of the solute become very small (negligible). So, the Volume of solution = 1L becomes equal to 1 kg. The weight of the solution can be treated as the weight of the solvent.

Q3. What is the difference between density and specific density?

Answer: Specific density of material calculated with respect to the water. So, it is unitless and absolute. Density calculated in g mL^-1 or other units.

Q4. Is mole fraction affected by temperature?

Answer: No, mole fraction is temperature independent. By measurement of solute & solvent masses, we can calculate the mole fraction of any components.

Q5. How do you change the concentration of a solution?

Answer: Sometimes, by modifying the quantity of solvent, a worker would need to modify the concentration of a solution. When a solvent is added to a solution, the concentration of solutes in the solution is diluted. Concentration involves the removal of solvent, which raises the concentration of solutes in the solution.

Q6. How do you prepare a solution of known concentration?

Answer: Solutions of known concentration can be prepared either by dissolving the known mass of the solvent solution and diluting it to the desired final volume or by diluting it to the desired final volume by diluting the acceptable volume of the more concentrated solution (the stock solution).