Higher Secondary Chemistry: mole fraction

Showing posts with label mole fraction. Show all posts

Thursday, August 27, 2020

Chapter 5.4 - Ideal Gas

In the previous section we completed a discussion on Avogadro law. We saw the details about STP also. In this section, we will see ideal gases

• A gas that follows Boyle's law, Charles' law and Avogadro law strictly is called an ideal gas
• From those three gas laws, we have three relations:
♦ At constant T and n, $\mathbf\small{\rm{V \propto \frac{1}{p}}}$ (Boyle's law)
♦ At constant p and n, V ∝ T (Charles' law)
♦ At constant p and T, V ∝ n (Avogadro law)
• Combining the three, we get: $\mathbf\small{\rm{V \propto \frac{nT}{p}}}$
■ Thus we get:
Eq.5.1: $\mathbf\small{\rm{V=R \frac{nT}{p}}}$
♦ R is the proportionality constant
✰ This R is called gas constant
♦ It is same for all gases
✰ So it is called Universal gas constant
• Eq.5.1 is called ideal gas equation

We want to find the value of R. It can be done in 4 steps:
1. To find R, we input all the known values into Eq.5.1
• Let us first write those known values:
• Consider one mole of a gas at STP
(i) Since there is only one mole, we can put n = 1
(ii) Since the gas is at STP, we can put:
♦ T = 273 K
(iii) Since the gas is at STP, we can put:
♦ p = 1 bar
(iv) We know that, one mole of a gas at STP will occupy 22.71 L
(v) Thus, R is the only unknown
2. We can substitute the known values in Eq.5.1 and obtain R
• But before that, we must convert all units into SI system
(i) n is a number. It can be written as such
(ii) T = 273 K is already in SI units
(iii) 1 bar is equal to 10⁵ Nm^-2
♦ So we must put p = 10⁵
(iv) 1 L is equal to 10^-3 m³
♦ So we must put V = 22.71 × 10^-3
3. Thus we get: $\mathbf\small{\rm{R=\frac{10^5 \times 22.71 \times 10^{-3}}{1 \times 273.15}}}$

= 8.314 Nm^-2 m³ K^-1 mol^-1

= 8.314 Pa m³ K^-1 mol^-1 (∵ 1 Nm^-2= 1 Pa)

= 8.314 × 10^-2bar L K^-1 mol^-1(∵ [Pa m³] = [(10^-5 bar) × (10³ L)] = 10^-2 bar L)

= 8.314 J K^-1 mol^-1. (∵ [Pa m³] = [(Nm^-2) × (m³)] = [N m] = J)
4. We see that, there are four possible SI units
• If any of the units written in (2) are changed, the value '8.314' will also change
♦ For example, if we put '32 ^oFahrenheit' in the place of '273 K', we will not get '8.314'
■ So we can write:
The value of R depends upon the units in which p, V and T are measured

• The Eq.5.1 is a relation between four variables
• Using that equation, we can describe the state of any gas
• So Eq.5.1 is called equation of state

Next, we will discuss about combined gas law. It can be written in 4 steps:
1. Let us rearrange the Eq.5.1 as: $\mathbf\small{\rm{\frac{pV}{T}=nR}}$
• If we perform experiments on the same sample of a gas, n will be a constant
• Then the 'right side as a whole' will become a constant
2. In such a situation, we measure the initial pressure, initial volume and initial temperature: p₁, V₁ and T₁
• These values will satisfy the rearranged equation in (1)
• So we get: $\mathbf\small{\rm{\frac{p_1V_1}{T_1}=nR}}$
3. Change the pressure to a new value p₂
• Measure the corresponding V₂ and T₂
• These values will also satisfy the rearranged equation in (1)
• So we get: $\mathbf\small{\rm{\frac{p_2V_2}{T_2}=nR}}$
4. Change the pressure to a new value p₃
• Measure the corresponding V₃ and T₃
• These values will also satisfy the rearranged equation in (1)
• So we get: $\mathbf\small{\rm{\frac{p_3V_3}{T_3}=nR}}$
so on . . .
4. All results are equal to nR
• So we can write: $\mathbf\small{\rm{\frac{p_1 V_1}{T_1}=\frac{p_2V_2}{T_2}=\frac{p_2V_2}{T_2}\;.\;.\;.\;so\;on\;.\;.\;.}}$
• From this we get:
Eq.5.2: $\mathbf\small{\rm{\frac{p_1 V_1}{T_1}=\frac{p_2V_2}{T_2}}}$
■ This equation is known as Combined gas law

We can bring density and molar mass also into the calculations. It can be written in 3 steps
1. Eq.5.1 can be rearranged as: $\mathbf\small{\rm{\frac{n}{V}=\frac{p}{RT}}}$
2. If m is the mass of the sample and M is the molar mass, we get:
• Number of moles (n) = $\mathbf\small{\rm{\frac{m}{M}}}$
• Substituting this in (1), we get: $\mathbf\small{\rm{\frac{m}{MV}=\frac{p}{RT}}}$
3. But $\mathbf\small{\rm{\frac{m}{V}}}$ is the density (d). So the result in (2) becomes: $\mathbf\small{\rm{\frac{d}{M}=\frac{p}{RT}}}$
• Rearranging this, we get:
Eq.5.3: $\mathbf\small{\rm{M=\frac{dRT}{p}}}$

Dalton’s law of partial pressures

• John Dalton did extensive research works on mixture of gases
• His findings were published as Dalton’s law of partial pressures
■ The law states that:
The total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases

• This can be explained in 8 steps:

1. In fig.5.19(a) below, a gas mixture is occupying a certain volume
• The 'occupied volume' is indicated by the grey rectangle

Fig.5.19

2. The gas mixture consists of three gases
• Gas 1, Gas 2 and Gas 3
♦ All molecules of Gas 1 are yellow in color
♦ All molecules of Gas 2 are red in color
♦ All molecules of Gas 3 are green in color
3. All molecules of Gas 1 are separated from the mixture
♦ Those separated yellow molecules are put in a second container
♦ This is shown in fig.b
4. All molecules of Gas 2 are separated from the mixture
♦ Those separated red molecules are put in a third container
♦ This is shown in fig.c
5. All molecules of Gas 3 are separated from the mixture
♦ Those separated green molecules are put in a fourth container
♦ This is shown in fig.d
6. The following three conditions must be satisfied:
(i) The gases in the mixture in fig.a must not react with each other
(ii) All the four containers must have the same volume
(iii) All the four containers must be at the same temperature
7. Relation between pressures:
♦ Let the pressure exerted by the mixture in fig.a be p_Total

♦ Let the pressure exerted by Gas 1 in fig.b be p₁

♦ Let the pressure exerted by Gas 2 in fig.c be p₂

♦ Let the pressure exerted by Gas 3 in fig.d be p₃

■ Then, according to Dalton’s law, we get:

p_Total = p₁+ p₂ + p₃ (at constant T, V)

■ This relation is applicable to any number of gases. So we can write:

Eq.5.4: p_Total = p₁+ p₂ + p₃ + . . . (at constant T, V)

• This is the mathematical form of Dalton's law of partial pressures

8. p₁, p₂, p₃. . . , which are the 'pressures exerted by individual gases in the mixture' are called partial pressures

Let us see a practical application of the law. It can be written in 5 steps:

1. Sometimes gases like CO₂, CH₃ etc., gets collected above water surface

• Then the 'gas collected above the water surface', will be a mixture of the original gas and water vapour

2. We are able to measure the 'pressure exerted by the mixture'

• But that measured pressure will include the pressure exerted by water vapour also

• We want the pressure exerted by the original gas alone

3. For that, we apply Dalton’s law of partial pressure

• We have: p_Total= p_{Water vapour}+ p_{Original gas}.

♦ So, if we subtract p_{Water vapour}from p_Total, we will get p_{Original gas}.

♦ That is., p_{Original gas} = p_Total- p_{Water vapour}

4. So our next task is to find p_{Water vapour}. It can be written in 5 steps:

(i) The air can hold a ‘certain maximum quantity’ of water vapour

■ When this maximum quantity is available in the air, we say that:

The air is saturated with water vapour

(ii) The saturation depends on temperature

♦ If the temperature is high, the air is able to hold more water vapour

✰ Consequently, more water vapour is required to make the air saturated

♦ If the temperature is low, the air is able to hold only less water vapour

✰ Consequently, less water vapour is sufficient to make the air saturated

(iii) We encounter different temperatures like 0 ^oC, 14 ^oC, 27 ^oC, 32 ^oC etc.,

• At each of those temperatures, air requires a ‘unique quantity of water vapour’ to become saturated

• Consequently, at each of those temperatures, there will be a ‘unique pressure’ exerted by the water vapour in the saturated air

■ This ‘unique pressure’ is called aqueous tension

(iv) Aqueous tension at various temperatures can be obtained from the data book

• Let us see some examples:

Example 1:

• Aqueous tension at 0 ^oC is 0.0060 bar

• That means, if air is saturated at 0 ^oC, the water vapour in that air will be exerting a pressure of 0.0060 bar

Example 2:

• Aqueous tension at 15 ^oC is 0.0168 bar

• That means, if air is saturated at 15 ^oC, the water vapour in that air will be exerting a pressure of 0.0168 bar

(v) Thus, once we know the temperature, we can obtain p_{Water vapour} from the data book

5. Now we can use the equation in (3) to obtain p_{Original gas}

• Note that in p_{Original gas}, there will not be even a single water molecule

♦ Because, p_{Water vapour} is already deducted

• So p_{Original gas}is also called p_{Dry gas}

■ Thus we get:

Eq.5.5: p_{Dry gas} = p_Total- p_{Water vapour}

Partial pressure in terms of mole fraction

This can be explained in 7 steps:

1. Consider fig.5.19 again. We have seen that:

• Pressure contributed by Gas 1 in fig.a = Pressure exerted by Gas 1 in fig.b

♦ But ‘pressure exerted by Gas 1 in fig.b’ = $\mathbf\small{\rm{p_1=\frac{n_1 RT}{V}}}$

✰ Where n₁ is the number of moles of Gas 1

• Pressure contributed by Gas 2 in fig.a = Pressure exerted by Gas 2 in fig.c

♦ But ‘pressure exerted by Gas 2 in fig.c’ = $\mathbf\small{\rm{p_2=\frac{n_2 RT}{V}}}$

✰ Where n₂ is the number of moles of Gas 1

• Pressure contributed by Gas 3 in fig.a = Pressure exerted by Gas 3 in fig.d

♦ But ‘pressure exerted by Gas 3 in fig.d’ = $\mathbf\small{\rm{p_3=\frac{n_3 RT}{V}}}$

✰ Where n₃ is the number of moles of Gas 3

2. So the total pressure p_Total will be given by:

p_Total = p₁+ p₂ + p₃= $\mathbf\small{\rm{\frac{n_1 RT}{V}+\frac{n_2 RT}{V}+\frac{n_3 RT}{V}}}$

$\mathbf\small{\rm{\Rightarrow\;p_{Total}=(n_1+n_2+n_3)\frac{ RT}{V}}}$

3. Let us divide p₁ by p_Total

• We get: $\mathbf\small{\rm{\frac{p_1}{p_{Total}}=\frac{n_1}{(n_1+n_2+n_3)}=\frac{n_1}{n}=X_1}}$

$\mathbf\small{\rm{\Rightarrow p_1=X_1 \times p_{Total}}}$

♦ Where:

✰ n = n₁+ n₂ + n₃.

✰ X₁ = mole fraction of the first gas

4. In a similar way, if we divide p₂ by p_Total, we will get:

$\mathbf\small{\rm{p_2=X_2 \times p_{Total}}}$

✰ X₂ = mole fraction of the second gas

5. In a similar way, if we divide p₃ by p_Total, we will get:

$\mathbf\small{\rm{p_3=X_3 \times p_{Total}}}$

✰ X₃ = mole fraction of the third gas

6. In general, we can write:

Eq.5.6: $\mathbf\small{\rm{p_i=X_i \times p_{Total}}}$

✰ X_i = mole fraction of the i^th gas

7. So if we know the total pressure, we can use Eq.5.6 to find the pressure exerted by individual gases

Now we will see some solved examples

Solved example 5.5 to 5.15

Solved example 5.16 to 5.24

• In the next section, we will see kinetic molecular theory of gases

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Monday, August 19, 2019

Chapter 1.5 - Molarity of Solutions

In the previous section, we saw the basics of stoichiometry. In this section we will see reactions in solutions

• On many occasions, reactions are carried out in solutions
• For example, in a reaction, a 'solution of NaOH in water' is one of the reactant
• In such cases, we use any one of the following four methods:
A. Mass percent or weight percent
B. Mole fraction
C. Molarity
D. Molality

We will see each of them in detail:
A. Mass percent
(i) In this method, we first write the masses:
• Mass of the solute (m₁)
• Mass of the solvent (m₂)
(ii) We then ask the question:
What is the 'contribution of the solute' towards making up the total mass (m₁+m₂) of the solution?
(iii) The answer can be written as follows:
$\mathbf\small{\left(\frac{m_1}{m_1+m_2}\right)^{th}}$ of the total mass is contributed by the solute
• This fraction when expressed as a percentage, gives the mass percent
• So we can write:
Eq.1.3: Mass percent = $\mathbf\small{\left(\frac{m_1}{m_1+m_2}\right)\times 100}$

An example:
5 g of a substance A is added to 20 grams of a solvent B. What is the mass percent of A?
Solution:
• We have:
Mass percent = $\mathbf\small{\left(\frac{m_1}{m_1+m_2}\right)\times 100}$
♦ Let mass of the substance A be m₁
♦ Let mass of the solvent B be m₂
• Substituting the values, we get:
Mass percent = $\mathbf\small{\left(\frac{5}{5+20}\right)\times 100=\left(\frac{1}{5}\right)\times 100=20\text{%}}$

B. Mole fraction
(i) In this method, we first write the 'number of moles':
• No. of moles of the solute (n₁)
• No. of moles of the solvent (n₂)
(ii) We then ask the question:
What is the 'contribution of the solute' towards making up the total number of moles (n₁+n₂)' in the solution?
(iii) The answer can be written as follows:
$\mathbf\small{\left(\frac{n_1}{n_1+n_2}\right)^{th}}$ of the 'total number of moles' is contributed by the solute
(iv) This fraction is called the mole fraction
• Note that, we do not need to convert this fraction into percentage form
• So we can write:
Eq.1.4: Mole fraction = $\mathbf\small{\left(\frac{n_1}{n_1+n_2}\right)}$

C. Molarity
• Detailed notes on molarity can be seen here.
• Based on those notes, we can now derive some interesting results:
• Consider the following situation:
(i) A solution of NaOH is available in a lab
(ii) This solution is named as: Solution A
(iii) The molarity of A is: 1 M
(iv) We want to take out a volume (V mL) from A
(v) Using this 'V mL', we want to make a new solution of NaOH
(vi) This new solution is named as: Solution B
(vii) The molarity of B should be 0.2 M
(viii) How much is 'V'?
• The above 8 points describe the situation
• But two points seem to be missing:
♦ What is the total volume of A?
♦ What is the total volume of B?
• For calculations, we can assume that:
♦ Volume of A = 1 L = 1000 mL
♦ Volume of B = 1 L = 1000 mL
• These assumptions will not affect the results
• Now we can write the steps for finding 'V':
1. 1000 mL of 1 M NaOH solution is kept in a beaker
• This is solution A
• Imagine a 'small cube of side 1 cm' within this solution
2. Volume of that cube = 1 cm³ = 1 cc = 1 mL
• There will be a total of 1000 such small cubes in A
• How many grams of NaOH will be present in that one small cube?
Answer:
(i) 1 M solution means: 1 mole is dissolved in 1000 mL
(ii) A has a volume of 1000 mL
(iii) So 1 mol of NaOH is present in A
(iv) That means 40 g of NaOH is present in A
(v) This 40 g is distributed uniformly among the 1000 small cubes of A
(vi) So one small cube will contain (⁴⁰⁄₁₀₀₀) = 0.04 g
3. Now consider B
(i) Molarity of B is to be: 0.2 M
(ii) 0.2 M solution means: 0.2 mol is dissolved in 1000 mL
(iii) B has a volume of 1000 mL
(iv) So 0.2 mol is to be present in B
(v) That means (0.2 × 40) = 8 g of NaOH is to be present in the 1000 mL of B
4. So we want 8 g of NaOH
(i) Our source is the solution A
(ii) We have seen that, each small cube in A will contain 0.04 g
(iii) Total number of cubes that will contain 8 g = (⁸⁄_0.04) = 200
(iv) So we must take out 200 cubes from A
(v) Each cube is 1 mL. So we can say:
We must take out 200 mL from A
5. We must take this 200 mL in a new beaker
• Add water gradually and bring up the final volume to 1000 mL
• This final 1000 mL solution will be having a molarity of 0.2 M

6. What if the original 1 M solution A has a volume of 1500 mL?

• If 1500 mL has a molarity of 1 M, then more than 40 g will be present in that 1550 mL
• How much more? Let us find out:
(i) We have: Molarity = (^mols⁄_{1000 mL})
(ii) 1 M means 1 mol is present in 1000 mL
(iii) So in 1 mL, there will be (¹⁄₁₀₀₀) = 0.001 mol
(iv) So in 1500 mL, there will be (1500 × 0.001) = 1.5 mols
(v) 1.5 mols = (1.5 × 40) = 60 g
(vi) So there should be 60 g in 1500 mL if it has to be a 1 M solution
7. In that case, the mass of NaOH contained in each small cube = (⁶⁰⁄₁₅₀₀) = 0.04 g
• To obtain 8 g, how many small cubes must be taken out from the 1500 mL?
• The answer is: (⁸⁄_0.04) = 200
• The same value obtained before
8. Thus it is clear that, the volume of the original solution does not matter. It is the molarity that matters
• Similarly, the 'initially assumed volume' of the new solution B also does not matter
• We have to bring up the 200 mL (taken out from A) to 1000 mL by adding water slowly
• So the final volume of B is 1000 mL

Based on the above discussion, we can derive a general formula:

Consider the following situation:
(i) A solution is available in a lab. It's volume is 1000 mL
(ii) This solution is named as: Solution A
(iii) The molarity of A is: M₁
(iv) We want to take out a volume (V mL) from A
(v) Using this 'V mL', we want to make a new solution
(vi) This new solution is named as: Solution B
(vii) The molarity of B is to be: M₂
(viii) How much is 'V'?
Solution:
1. Let there be m₁ grams dissolved in the 1000 mL of A
2. Each 'small cube of 1 cm side' in A will contain $\mathbf\small{\left(\frac{m_1}{1000}\right)}$ grams
3. Now consider B
• Molarity of B is to be: M₂
• Let m₂ be the mass to be taken out from A
4. Then number of cubes to be taken out = $\mathbf\small{\left(\frac{m_2}{\frac{m_1}{1000}}\right)=\frac{1000m_2}{m_1}}$
• Each cube has a volume of 1 mL
5. So we can write:
Volume to be taken out from the original solution = $\mathbf\small{\left(\frac{1000\,m_2}{m_1}\right)\text{mL}}$
6. Consider the ratio $\mathbf\small{\left(\frac{m_2}{m_1}\right)}$
• Both m₂ and m₁ are in grams
• Dividing both numerator and denominator by 1000 mL, we get:
$\mathbf\small{\left(\frac{m_2}{m_1}\right)}$ = $\mathbf\small{\left[\frac{\left(\frac{m_2}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}{\left(\frac{m_1}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}\right]}$
• But $\mathbf\small{\left(\frac{m_2}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}$ = The molarity M₂
• Similarly, $\mathbf\small{\left(\frac{m_1}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}$ = The molarity M₁
• So the ratio $\mathbf\small{\left(\frac{m_2}{m_1}\right)}$ can be written as $\mathbf\small{\left(\frac{M_2}{M_1}\right)}$
7. So the result in (5) will become:

Eq.1.5:

Volume to be taken out from the original solution = $\mathbf\small{\left(\frac{1000\,M_2}{M_1}\right)\text{mL}}$

Let us apply this equation to our present case:

1. Molarity of the original solution of NaOH= 1 M
♦ This is M₁
2. Molarity of the new solution = 0.2 M
♦ This is M₂
3. So volume to be taken out from the original solution = $\mathbf\small{\left(\frac{1000\,M_2}{M_1}\right)=\left(\frac{1000\times 0.2}{1.0}\right)=200\,\text{mL}}$
4. This 200 mL should be gradually brought up to 1000 mL by adding water
• The final 1000 mL solution thus obtained, will have a molarity of 0.2 M

Solved example 1.20
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution
Solution:
1. Assume that, the 250 mL is taken out from a solution A of volume 1000 mL
2. Then we can write:
250 mL of A has 4 g
3. So 1 mL of A has (⁴⁄₂₅₀) g
4. So 1000 mL of A has [(⁴⁄₂₅₀) × 1000] = 16 g
5. So 16 g is dissolved in 1000 mL
6. We can write:
4 g dissolved in 250 mL
is equivalent to
16 g dissolved in 1000 mL
7. One mole of NaOH = 40 g
• So 1 g = (¹⁄₄₀) mole
• So 16 g = [16 × (¹⁄₄₀)] = 0.4 mole
8. We can write:
• 0.4 mole is dissolved in 1000 mL
• But 'no. of moles dissolved in 1000 mL' is the molarity
9. So we get:
4 g dissolved in 250 mL
is equivalent to
a molarity of 0.4 M

Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent
• When the temperature increases, the volume also increases
♦ The number of moles which is in the numerator, remains the same
♦ The volume which is in the denominator increases
♦ So molarity decreases
• When the temperature decreases, the volume also decreases
♦ The number of moles which is in the numerator, remains the same
♦ The volume which is in the denominator decreases
♦ So molarity increases

D. Molality
In this method we first write two items:
(i) The number of moles of the solute
(ii) The mass of the solvent in kg
• Their ratio is called molality. It's symbol is m
• So we get:
Eq.1.6: Molality (m) = $\mathbf\small{\frac{\text{No. of moles of solute}}{\text{mass of solvent in kg}}}$
• So in effect, m is the 'number of moles of the solute' present in 'each 1 kg' of the solvent

Solved example 1.21
The density of 3 M solution of NaCl is 1.25 g mL^-1. Calculate the molality of the solution
Solution:
1. Consider 1000 mL volume of the given solution
• What ever be the volume, the density will not change
• So the 1000 mL of the given solution will have a density of 1.25 g mL^-1.
2. We have: $\mathbf\small{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$
$\mathbf\small{\Rightarrow\text{1.25 (g mL)}^{-1}=\frac{\text{Total mass of the solution}}{\text{1000 (mL)}}}$
$\mathbf\small{\Rightarrow \text{Total mass of the solution}= \text{1.25 (g mL)}^{-1}\times 1000\,\text{(mL)}= 1250\, \text{g}}$
3. This 1250 g is the total mass of the solution
• So we can write:
1250 g = Mass of the NaCl + Mass of the solvent
4. Molarity of the solution is given as 3 M
• So every 1000 mL of the solution will contain 3 moles of NaCl
5. Three moles of NaCl = (3 × 58.5) = 175.5 g
• So from the result in (3), we get:
1250 = 175.5 + Mass of the solvent
⇒ Mass of the solvent = (1250-175.5) = 1074.5 g = 1.0745 kg
6. So, for every 1.0745 kg, there are 3 moles of NaCl
• Thus we get:
Molatity = No. of moles per kg = $\mathbf\small{\frac{\text{3 (mole)}}{\text{1.0745 (kg)}}=2.79\, \text{mol kg}^{-1}}$

We have completed a discussion on the basics of the four methods. Now we will see some more solved examples

Solved example 1.22
Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^-1.
Solution:
1. We want a 0.375 M solution of sodium acetate
• So 1000 mL of that solution should contain 0.375 moles
2. One mole of sodium acetate is 82.0245 g
• So 0.375 mol = (0.375× 82.0245) = 30.7592 g
3. This 30.7592 g of sodium acetate is to be dissolved in 1000 mL
• So for 500 mL, (^30.7592⁄₂) = 15.3796 g of sodium acetate will be sufficient

Solved example 1.23
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL^-1 and the mass per cent of nitric acid in it being 69%.
Solution:
1. Consider 1000 mL volume of the sample
• What ever be the volume, the density will not change
• So the 1000 mL of the sample will have a density of 1.41 g mL^-1.
2. We have: $\mathbf\small{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$
$\mathbf\small{\Rightarrow\text{1.41 (g mL)}^{-1}=\frac{\text{Total mass of the solution}}{\text{1000 (mL)}}}$
$\mathbf\small{\Rightarrow \text{Total mass of the solution}= \text{1.41 (g mL)}^{-1}\times 1000\,\text{(mL)}= 1410\, \text{g}}$
3. This 1410 g is the total mass of the solution
• So we can write:
1410 g = Mass of the nitric acid + Mass of the solvent
4. Given that, mass percent of nitric acid is 69%
• That means, 69% of 1410 g is the 'mass of nitric acid'
• So we get:
Mass of nitric acid in 1000 mL solution = (1410 × 0.69) = 972.9 g
5. One mol of nitric acid (HNO₃) = (1 × 1.008) + (1 × 14) + (3 × 16) = 63.01 g
• So 972.9 g = (^972.9⁄_63.01) = 15.44 mols
6. So we get:
• 15.44 mols will be present in 1 liter
• So molarity of the solution is 15.44 M

Solved example 1.24
What is the molarity of H₂SO₄ solution that has a density of 1.84 g/cc at 35^o c and contains 98% by weight?
Solution:
1. Consider 1000 mL volume of the given solution
• What ever be the volume, the density will not change
• So the 1000 mL of the given solution will have a density of 1.84 g mL^-1
(∵ 1 cc = 1 mL)
2. We have: $\mathbf\small{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$
$\mathbf\small{\Rightarrow\text{1.84 (g mL)}^{-1}=\frac{\text{Total mass of the solution}}{\text{1000 (mL)}}}$
$\mathbf\small{\Rightarrow \text{Total mass of the solution}= \text{1.84 (g mL)}^{-1}\times 1000\,\text{(mL)}= 1840\, \text{g}}$
3. This 1840 g is the total mass of the solution
• Given that 98% of the total mass is H₂SO₄
• So we can write:
• Mass of H₂SO₄ = (1840 × 0.98)
4. Molar mass of H₂SO₄= (2 ×1) + (1 × 32) + (4 × 16) = 98 g
• So number of moles in the 1000 mL = $\mathbf\small{\frac{1840 \times 0.98}{98}}$ = 18.40 mols
5. But 'number of mols in 1000 mL' is molarity
• So we can write:
Molarity of the given solution is 18.4 M

In the next section, we will see a few more solved examples