Molarity: Definition, Concept of Dilution, Practice Problems and FAQs

We often drink lemonade and just by tasting it we can tell what particular flavour is missing and which ingredient is in less or more amount. So, to get the best flavour, all the ingredients should be in the proper quantity in the lemonade.

To prepare the best lemonade solution, ingredients like sugar, salt, lemon juice and water should be in the exact same quantity. Ingredients like sugar, salt and lemon juice are the solute in the lemonade solution and water is the solvent. The amount of solute in a solution is known as the concentration of the solution. One such method to express the concentration of the solution is Molarity.

Let’s learn more about the concept of Molarity.

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Table of Content

Relationship Between Moles & Mass
What is Molarity?
Concept of Dilution
Mixing of the Solutions
Practice Problems
Frequently Asked Questions - FAQs

Relationship Between Moles & Mass

The amount of material that contains the same number of essential components as the number of atoms found in a pure sample of carbon weighing exactly 12 g is known as a mole. Total 6.0221023 entities make up the value of 1 mole of any substance. Hence, The mass of 6.0221023 particles of any substance is termed as the molar mass of the substance.

$N u m b e r o f m o l e s = \frac{m a s s o f s u b s t a n c e (g)}{M o l a r m a s s (g m o l^{- 1})}$

What is Molarity?

A binary solution is made up of two components that are solvent and solute. Molarity is the concentration term which is used to define the concentration of a solution. In simple terms, molarity can be defined as the concentration of solute in a given volume of solution.
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When one gram of solute dissolves in one litre of solution, the solution has a molarity of one. Since the solvent and solute combine to form a solution, the total volume of the solution is measured.

To simplify, let’s understand it mathematically,

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

Molarity is denoted by M and its unit is mol L-1.

For example:

Let’s s suppose, the molarity of an aqueous solution of sugar is 2 M.

This means that 2 mol sugar is dissolved in 1 L of the solution.

For example:

Find the molarity of 18.25 g HCl dissolved in 250 mL solution.

Molecular mass of HCl=(1atomic mass of H)+(1atomic mass of Cl)

Molecular mass of HCl=36.5 u

Molar mass of HCl=36.5 g mol-1

Given mass of HCl=18.25 g

Volume of the solution = 250 mL

$M o l a r i t y = \frac{m a s s o f s o l u t e (g) \times 1000}{M o l a r m a s s o f t h e s o l u t e (g m o l^{- 1}) \times v o l u m e o f s o l u t i o n (m L)}$

$M o l a r i t y = \frac{18.25 g \times 1000}{36.5 g m o l^{- 1} \times 250} = 2 m o l L^{- 1} o r 2 M$

Hence, the molarity of 18.25 g of HCl in 250 mL of solution is 2M.

Concept of Dilution

In simple terms, the dilution is changing the volume of the solution by keeping the amount of the solute constant and adding more solvent to it which ultimately reduces the concentration of the solution.

For example, 50 g the salt is dissolved in 500 mL of water. Now to dilute the salt solution, another 500 mL water is added. Now, the question here is what would be the molarity of the new solution after dilution?

Let’s understand it and derive a formula to calculate the molarity after dilution.

So, the initial molarity $(M_{1}) = \frac{m a s s o f s o l u t e (g) \times 1000}{M o l a r m a s s o f t h e s o l u t e (g m o l^{- 1}) \times v o l u m e o f s o l u t i o n (m L)}$

$(M_{1}) = \frac{50 g \times 1000}{58.5 g m o l^{- 1} \times 500} = 1.709 M$

The final molarity $(M_{2}) = \frac{m a s s o f s o l u t e (g) \times 1000}{M o l a r m a s s o f t h e s o l u t e (g m o l^{- 1}) \times v o l u m e o f s o l u t i o n (m L)}$

$(M_{2}) = \frac{50 g \times 1000}{58.5 g m o l^{- 1} \times 1000} = 0.854 M$ (∵ the total volume of the solution now is 1000 mL)

Deriving a formula to calculate the molarity of the solution when it diluted from V1 to V2:

Initial molarity = M1 (known), Final molarity = M2

Initial volume = V1, diluted to volume = V2

Number of moles of solute (n) is kept constant.

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

$M_{1} = \frac{n}{V_{1}}$

$n = M_{1} \times V_{1} . . . . (i)$

$M_{2} = \frac{n}{V_{2}}$

$n = M_{2} \times V_{2} . . . . (i i)$

Comparing equation (i) & (ii);

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

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

For example: 40 g of NaOH is dissolved in 250 mL solution. What will be the concentration of the solution if the volume of solution is changed to 1 L?

Molar mass of NaOH=40 g mol-1

Given mass of NaOH=40 g

Initial volume of the solution (V1) = 250 mL

$M o l a r i t y (M_{1}) = \frac{m a s s o f N a O H (g) \times 1000}{M o l a r m a s s o f N a O H (g m o l^{- 1}) \times v o l u m e o f s o l u t i o n (m L)}$

$M o l a r i t y (M_{1}) = \frac{40 g \times 1000}{40 g m o l^{- 1} \times 250} = 4 M$

Final concentration;

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

$M_{2} = \frac{4 M \times 250 m L}{1000 m L} = 1 M$

Mixing of the Solutions

Calculating the molarity of the mixture of solutions with one solution having molarity (M1) & volume (V1) mixed with a solution having the same solute with molarity (M2) & volume (V2)

Let’s consider;

Molarity of the Solution 1 = M1 (known), Molarity of the Solution 2 = M2

Volume of the Solution 1 = V1, Volume of the Solution 2 = V2

Final Molarity of the Solution = MF, Final Volume of the Solution = V1+V2

Number of moles of solute (n) will be constant before and after the mixing of the solutions.

M1=n1V1

⇒ n=M1V1....(i)

M2=n2V2

⇒ n=M2V2....(ii)

MF=nFV1+ V2

⇒ nF=MF(V1+ V2)....(iii)

Now, nF=n1+n2

Substituting the values of nF , n1 & n2, we get;

⇒ MF(V1+ V2)= M1V1+M2V2

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

For Example: If two solutions, 4M, 200 ml and 6M, 400 ml, are combined. Determine the molarity of the resulting solution.

M1=4M

M2=6M

V1=200 ml

V2=400 ml

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

$M_{F} = \frac{4 M \times 200 m l + 6 M \times 400 m l}{200 m l + 400 m l} = \frac{(800 + 2400)}{600} = 5.33 M$

Practice Problems

Q1. Find the molarity of 20 g NaOH dissolved in 250 ml solution. Given: Atomic mass of Na = 23 u, H = 1 u, O = 16 u

Answer: (B)

Molecular mass of NaOH=(1atomic mass of Na)+(1atomic mass of O) )+(1atomic mass of H)

Molecular mass of NaOH=40 u

Molar mass of NaOH=40 g mol-1

Given mass of NaOH=20 g

Volume of the solution (V) = 250 mL

$M o l a r i t y (M) = \frac{m a s s o f N a O H (g) \times 1000}{M o l a r m a s s o f N a O H (g m o l^{- 1}) \times v o l u m e o f s o l u t i o n (m L)}$

$M o l a r i t y (M) = \frac{20 g \times 1000}{40 g m o l^{- 1} \times 250} = 2 M$

Q2. Find the molarity of 4.9g H2SO4 present in 500 cm3 solution.

1M
2M
10M
0.1M

Answer: (D)

Molecular mass of H2SO4=(2atomic mass of H)+(1atomic mass of S) )+(4atomic mass of O)

Molecular mass of H2SO4=98 u

Molar mass of H2SO4=98 g mol-1

Given mass of H2SO4=4.9 g

Now we have to convert 500 cm3 into mL.

500 cm3=500 mL

$M o l a r i t y (M) = \frac{m a s s o f H_{2} S O_{4} (g) \times 1000}{M o l a r m a s s o f H_{2} S O_{4} (g m o l^{- 1}) \times v o l u m e o f s o l u t i o n (m L)}$

$M o l a r i t y (M) = \frac{4.9 g \times 1000}{98 g m o l^{- 1} \times 500} = 0.1 M$

Q3. How much more water is needed to change the concentration of 3M Na2CO3 in 100 mL to 2M?

35 mL
200 mL
50 mL
150 mL

Answer: (C)

M1=3M

M2=2M

V1=100 mL

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

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

$V_{2} = \frac{3 M \times 100 m L}{2 M} = 150 m L$

So the new volume is 150 mL.

As initial volume was 100 mL.

Hence, 150 mL-100 mL = 50 mL more water is needed to change the concentration of the solution from 3M to 2M.

Q4. The two HCl solutions with 2.0 M and 0.5 M, are mixed. The volumes of the solutions that were taken to make a 1.5 M HCl solution are:

30 mL, 60 mL
100 mL, 90 mL
70 mL, 30 mL
60 mL, 30mL

Answer: (D)

M1=2M

M2=0.5M

V1=? mL

V2=? mL

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

$1.5 M = \frac{2 V_{1} + 0.5 V_{2}}{V_{1} + V_{2}}$

${1.5 V}_{1} + {1.5 V}_{2} = 2 V_{1} + 0.5 V_{2}$

0.5 V1= V2

$\frac{V_{1}}{V_{2}} = \frac{2}{1}$

From the above mentioned options only option D satisfies the correct ratio.

$\frac{60 m l}{30 m l} = \frac{2}{1}$

Hence, option A is the correct choice.

Frequently Asked Questions - FAQs

Q1: What will happen to the value of molarity if temperature changes?
Answer: Molarity is a concentration term which is used to define the concentration of a solution. In simple terms, molarity can be defined as the concentration of solute in a given volume of solution.

$M o l a r i t y = \frac{N u m b e r o f m o l e s o f s o l u t e}{v o l u m e o f s o l u t i o n i n l i t r e s}$

When the temperature of the solution increases the molecules gain kinetic energy and start moving. This increases the volume of the solution. As molarity is calculated in terms of the volume of the solution. Hence, altering the value of temperature will change the value of molarity. With the rise in temperature, molecules gain kinetic energy and start moving, which increases the volume of the solution. As molarity is inversely proportional to the volume of the solution, hence, molarity decreases when temperature increases.

Q2: Are molarity and molality equivalent terms?
Answer: No, even though there’s only one letter that sets them apart, still they both carry different meanings. Obviously, both molarity and molality represent the concentration of a solution. But their way of representation is different. Molarity is expressed in terms of the volume of the solution whereas molality is expressed in terms of the mass of the solvent. Molarity gets easily affected by the temperature change whereas molality remains unaffected by changing temperature.

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

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

Q3: Can we measure the molarity of a solution, where the solute is partially dissolved?
Answer: No, molarity defines the concentration of a solute in a given volume of solution. If the solute is partially dissolved then, we have to find out how much solute is dissolved in it and how much is undissolved, which is quite a complicated task. Hence, it is advisable to calculate the molarity of any solution after the complete dissolution of the solute.

Q4: Is molarity an extensive property or intensive property?
Answer: An intensive property is one that is independent of a system's or substance's mass. Whereas extensive property is dependent on the mass of the system or substance. Molarity is an intensive property. You can take the example of juice. If you take out some juice from a glass full of juice, will it change the taste of the juice? No, it won’t change. So, the concentration remains the same. Hence, molarity is an intensive property.