Important Questions Class 12 Chemistry Chapter 2 Solutions 5 Marks Questions

CBSE Class 12 Chemistry Chapter 2 Important Questions – Free PDF Download

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CBSE Class 12 Chemistry Important Questions Chapter 2 – Solutions

5 Marks Questions

1. The vapour pressure of at  is 854 mm Hg .A solution of 2.0g sulphur in 100g  of  has a vapour pressure of 848.9 mm Hg .Calculate the formula
 of sulphur molecule.
Ans. = 854 mm, ,



= 
= 
= 254.5  g/mol.
Let the formula = Sx

X= 
= 7.95
= 8.
= Formula = 
2. Calculate the mass percentage of benzene () and carbon tetrachloride () if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Ans. Mass percentage of 

= 15.28%
Mass percentage of =

=84.72%
Alternatively,
Mass percentage of = (100 – 15.28)%
= 84.72%
3. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Ans. Let the total mass of the solution be 100 g and the mass of benzene be 30 g.
∴Mass of carbon tetrachloride = (100 – 30)g
= 70 g
Molar mass of benzene () = (612 + 6) 
= 78 
∴Number of moles of 
= 0.3846 mol
Molar mass of carbon tetrachloride ( ) = 112 + 4355
= 154 
∴Number of moles of 
= 0.4545 mol
Thus, the mole fraction of is given as:

 
= 0.458
4. Calculate the molarity of each of the following solutions: (a) 30 g of .  in 4.3 L of solution (b) 30 mL of 0.5 M diluted to 500 mL.
Ans. Molarity is given by:

(a) Molar mass of  = 59 + 2 (14 + 3 16) + 618
= 291 
Therefore, Moles of 
= 0.103 mol
Therefore, molarity 
= 0.023 M
(b) Number of moles present in 1000 mL of 0.5 M 
∴Number of moles present in 30 mL of 0.5 M 
= 0.015 mol
Therefore, molarity
= 0.03 M
5. Calculate (a)molality (b)molarity and (c)mole fraction of KI if the density of 20% (mass/mass) aqueous KI is.
Ans. (a) Molar mass of KI = 39 + 127 = 166 
20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.
That is,
20 g of KI is present in (100 – 20) g of water = 80 g of water
Therefore, molality of the solution =

= 1.506 m
= 1.51 m (approximately)
(b) It is given that the density of the solution = 
Therefore, Volume of 100 g solution = 
=
= 83.19 mL
= 
Therefore, molarity of the solution = 
= 1.45 M
(c) Moles of KI = 
Moles of water = 
Therefore, mole fraction of KI = 
= 
= 0.0263
6. , a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of  in water at STP is 0.195 m, calculate Henry’s law constant.
Ans. It is given that the solubility of in water at STP is 0.195 m, i.e., 0.195 mol of is dissolved in 1000 g of water.
Moles of water = 
= 55.56 mol
∴Mole fraction of , x = 
= 
= 0.0035
At STP, pressure (p) = 0.987 bar
According to Henry’s law:
p= KHx

= 282 bar
7.  Henry’s law constant for  in water is 1.67 x 108 Pa at 298 K. Calculate the quantity of in 500 mL of soda water when packed under 2.5 atm pressure at 298 K.
Ans. It is given that:
KH = 
= 2.5 atm = 
= 
According to Henry’s law:



= 0.00152
We can write, 
[Since, is negligible as compared to]
In 500 mL of soda water, the volume of water = 500 mL
[Neglecting the amount of soda present]
We can write:
500 mL of water = 500 g of water
= mol of water
= 27.78 mol of water
Now,


Hence, quantity of in 500 mL of soda water = 
= 1.848 g
8.  The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Ans. It is given that:
 = 450 mm of Hg
= 700 mm of Hg
P total= 600 mm of Hg
From Raoult’s law, we have:


Therefore, total pressure, 






Therefore, 
= 1 – 0.4
= 0.6
Now, 
= 
= 180 mm of Hg

= 
= 420 mm of Hg
Now, in the vapour phase:
Mole fraction of liquid A = 


= 0.30
And, mole fraction of liquid B = 1 – 0.30
9.  Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea () is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Ans. It is given that vapour pressure of water, of Hg
Weight of water taken, 
Weight of urea taken, 
Molecular weight of water, 
Molecular weight of urea, 
Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1.
Now, from Raoult’s law, we have:






Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.
10. Boiling point of water at 750 mm Hg is . How much sucrose is to be added to 500 g of water such that it boils at . Molal elevation constant for water is.
Ans. Here, elevation of boiling point = (100 + 273) – (99.63 + 273)
= 0.37 K
Mass of water, 
Molar mass of sucrose ( ), 
= 
Molal elevation constant, Kb = 
We know that:



= 121.67 g (approximately)
Hence, 121.67 g of sucrose is to be added.
11. Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Ans. (i) Mole fraction:
The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture.
i.e.,
Mole fraction of a component = 
Mole fraction is denoted by ‘x’.
If in a binary solution, the number of moles of the solute and the solvent are nA and nB respectively, then the mole fraction of the solute in the solution is given by,

Similarly, the mole fraction of the solvent in the solution is given as:

(ii) Molality
Molality (m) is defined as the number of moles of the solute per kilogram of the solvent. It is expressed as:
Molality (m) = 
(iii) Molarity
Molarity (M) is defined as the number of moles of the solute dissolved in one Litre of the solution.
It is expressed as:
Molarity (M) = 
(iv) Mass percentage:
The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100 g of the solution. It is expressed as:
Mass % of a component = 
12. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is?
Molarity of solution = 
= 16.23 M
Ans. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution.
Molar mass of nitric acid 
Then, number of moles of 
= 1.079 mol
Given,
Density of solution = 
Therefore, Volume of 100 g solution = 
= 66.49 mL
= 
Molarity of solution = 
= 16.23 M
13. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is  , then what shall be the molarity of the solution?
Ans. 10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 – 10) g = 90 g of water.
Molar mass of glucose 
Then, number of moles of glucose = 
= 0.056 mol
Molality of solution =  = 0.62 m
Number of moles of water = 
= 5 mol
Mole fraction of glucose 
= 0.011
And, mole fraction of water 
= 1 – 0.011
= 0.989
If the density of the solution is , then the volume of the 100 g solution can be given as:

= 83.33 mL
= 
//www.schoollamp.com/images/ncert-solutions/chemistry+solutions+cbse+14167336860342.gifMolarity of the solution = 
= 0.67 M
14. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of  and containing equimolar amounts of both?
Ans. Let the amount of in the mixture be x g.
Then, the amount of in the mixture is (1 – x) g.
Molar mass of 
= 
//www.schoollamp.com/images/ncert-solutions/chemistry+solutions+cbse+1416733688595.gifNumber of moles = 
Molar mass of 
= 
Number of moles of  = 
According to the question,

⇒ 84x = 106 – 106x
⇒ 190x = 106
⇒ x = 0.5579
Therefore, number of moles of = 
= 0.0053 mol
And, number of moles of = 
= 0.0053 mol
HCl reacts with and according to the following equation.


1 mol of  reacts with 2 mol of HCl.
Therefore, 0.0053 mol of reacts with = 0.0106 mol.
Similarly, 1 mol of  reacts with 1 mol of HCl.
Therefore, 0.0053 mol of reacts with 0.0053 mol of HCl.
Total moles of HCl required = (0.0106 + 0.0053) mol
= 0.0159 mol
In 0.1 M of HCl,
0.1 mol of HCl is preset in 1000 mL of the solution.
Therefore, 0.0159 mol of HCl is present in 
= 159 mL of the solution
Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of and , containing equimolar amounts of both.
15. An antifreeze solution is prepared from 222.6 g of ethylene glycol () and 200 g of water. Calculate the molality of the solution. If the density of the solution is, then what shall be the molarity of the solution?
Ans. Molar mass of ethylene glycol 
= 
Number of moles of ethylene glycol 
= 3.59 mol
Therefore, molality of the solution =
= 17.95 m
Total mass of the solution = (222.6 + 200) g
= 422.6 g
Given,
Density of the solution = 1.072 g mL – 1
Therefore, Volume of the solution = 
= 394.22 mL
= 
Molarity of the solution = 
= 9.11 M
16. State Henry’s law and mention some important applications?
Ans. Henry’s law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas, then Henry’s law can be expressed as:

Where,
is Henry’s law constant
Some important applications of Henry’s law are mentioned below.
(i) Bottles are sealed under high pressure to increase the solubility of in soft drinks and soda water.
(ii) Henry’s law states that the solubility of gases increases with an increase in pressure. Therefore, when a scuba diver dives deep into the sea, the increased sea pressure causes the nitrogen present in air to dissolve in his blood in great amounts. As a result, when he comes back to the surface, the solubility of nitrogen again decreases and the dissolved gas is released, leading of the formation of nitrogen bubbles in the blood. This results in the blockage of capillaries and leads to a medical condition known as ‘bends’ or ‘decompression sickness’.
Hence, the oxygen tanks used by scuba divers are filled with air and diluted with helium to avoid bends.
(iii) The concentration of oxygen is low in the blood and tissues of people living at high altitudes such as climbers. This is because at high altitudes, partial pressure of oxygen is less than that at ground level. Low-blood oxygen causes climbers to become weak and disables them from thinking clearly. These are symptoms of anoxia
17. The partal pressure of ethane over a solution containing of ethane is 1 bar. If the solution contains  of ethane, then what shall be the partial pressure of the gas?
Ans. Molar mass of ethane 
= 
Number of moles present in of ethane = 
= 
Let the number of moles of the solvent be x.
According to Henry’s law,

1 bar 
1 bar = 

18. What is meant by positive and negative deviations from Raoult’s law and how is the sign of  related to positive and negative deviations from Raoult’s law?
Ans. According to Raoult’s law, the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult’s law (non-ideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.

Vapour pressure of a two-component solution showing positive deviation from Raoult’s law

Vapour pressure of a two-component solution showing negative deviation from Raoult’s law
In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero.
ΔsolH= 0
In the case of solutions showing positive deviations, absorption of heat takes place.
Therefore, = Positive
In the case of solutions showing negative deviations, evolution of heat takes place.
Therefore, = Negative
19. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Ans. Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar
Vapour pressure of pure water at normal boiling point 
Mass of solute,  = 2 g
Mass of solvent (water), = 98 g
Molar mass of solvent (water), = 
According to Raoult’s law,




= 
Hence, the molar mass of the solute is .
20. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Ans. Vapour pressure of heptanes 
Vapour pressure of octane = 46.8 kPa
We know that,
Molar mass of heptane 
= 
Therefore, Number of moles of heptane = 
= 0.26 mol
Molar mass of octane 
= 
Therefore, Number of moles of octane = 
= 0.31 mol
Mole fraction of heptane, 
= 0.456
And, mole fraction of octane, = 1 – 0.456
= 0.544
Now, partial pressure of heptane, 
= 
= 47.97 kPa
Partial pressure of octane, 
= 
= 25.46 kPa
Hence, vapour pressure of solution, 
= 47.97 + 25.46
= 73.43 kPa
21. The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Ans. 1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water = 
Therefore, Number of moles present in 1000 g of water = 
= 55.56 mol
Therefore, mole fraction of the solute in the solution is
.
It is given that,
Vapour pressure of water,  = 12.3 kPa
Applying the relation, 

⇒ 12.3 – = 0.2177
⇒ = 12.0823
= 12.08 kPa (approximately)
Hence, the vapour pressure of the solution is 12.08 kPa.
22. Calculate the mass of a non-volatile solute (molar mass 40 g) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Ans. Let the vapour pressure of pure octane be
Then, the vapour pressure of the octane after dissolving the non-volatile solute is 
Molar mass of solute, 
Mass of octane, = 114 g
Molar mass of octane, 
= 
Applying the relation,





Hence, the required mass of the solute is 8 g.
23. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
1. molar mass of the solute
2. vapour pressure of water at 298 K.
Ans. (i) Let, the molar mass of the solute be 
Now, the no. of moles of solvent (water), 
And, the no. of moles of solute, 
p1=2.8kpap1=2.8kpa
Applying the relation:








After the addition of 18 g of water:


Again, applying the relation:








Dividing equation (i) by (ii), we have:



87M + 435 = 84 M +504
3M = 69
M = 23u
Therefore, the molar mass of the solute is.
(ii) Putting the value of ‘M’ in equation (i), we have:



Hence, the vapour pressure of water at 298 K is 3.53 kPa.
24. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Ans. Here, = (273.15 – 271) K
= 2.15 K
Molar mass of sugar 
= 
5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 – 5)g = 95 g of water.
Now, number of moles of cane sugar = 
= 0.0146 mol
Therefore, molality of the solution, 
= 
Applying the relation,



= 
Molar of glucose 
= 
5% glucose in water means 5 g of glucose is present in (100 – 5) g = 95 g of water.
Therefore, Number of moles of glucose = 
= 0.0278 mol
Therefore, molality of the solution, 
= 
Applying the relation,

= 
= 4.09 K (approximately)
Hence, the freezing point of 5% glucose solution is (273.15 – 4.09) K= 269.06 K.
25. Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene, 1 g of AB2 lowers the freezing point by 2.3 Kwhereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is . Calculate atomic masses of A and B.
Ans. We know that,

Then, 
= 

= 
Now, we have the molar masses of and as andrespectively.
Let the atomic masses of A and B be x and y respectively.
Now, we can write:

Subtracting equation (i) from (ii), we have
2y = 85.28
⇒ y = 42.64
Putting the value of ‘y‘ in equation (1), we have

⇒ x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.
26. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol (ii) toluene (iii) formic acid
(iv) ethylene glycol (v) chloroform (vi) pentanol.
Ans. (i) Phenol has the polar group -OH and non-polar group. Thus, phenol is partially soluble in water.
(ii) Toluene has no polar groups. Thus, toluene is insoluble in water.
(iii) Formic acid (HCOOH) has the polar group -OH and can form H-bond with water. Thus, formic acid is highly soluble in water.
(iv) Ethylene glycol has polar -OH group and can form H-bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water.
(vi) Pentanol   has polar -OH group, but it also contains a very bulky non-polar ­­­ group. Thus, pentanol is partially soluble in water.
27. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Ans.



Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:
Acetic acid < trichloroacetic acid < trifluoroacetic acid
28. Calculate the depression in the freezing point of water when 10 g of is added to 250 g of water. .
Ans. Molar mass of 
= 
Therefore, No. of moles present in 10 g of 
= 0.0816 mol
It is given that 10 g of is added to 250 g of water.
Therefore, Molality of the solution, 
= 0.3264 
Let a be the degree of dissociation of 
undergoes dissociation according to the following equation:
Initial conc. At equilibrium = 


Since α is very small with respect to 1, 1 – α -1

Now, 


= 0.0655
Again,
Initial moles at equilibrium = 
Total moles of equilibrium = 
= 
Therefore, 

= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:


= 0.65 K
29. 19.5 g of is dissolved in 500 g of water. The depression in the freezing point of water observed is . Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Ans. It is given that:




We know that: 


Therefore, observed molar mass of 
The calculated molar mass of is:


Therefore, van’t Hoff factor, 

= 1.0753
Let αbe the degree of dissociation of 
Initial conc. At equilibrium = Total = 



= 1.0753 – 1
= 0.0753
Now, the value of is given as:



Taking the volume of the solution as 500 mL, we have the concentration:

= 0.5 M
Therefore, 


= 0.00307 (approximately)

30. Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in 450 g of water.
Ans. Vapour pressure of water, = 17.535 mm of Hg
Mass of glucose, 
Mass of water, 
We know that,
Molar mass of glucose 
= 180 g 
Molar mass of water, 
Then, number of moles of glucose, 
= 0.139 mol
And, number of moles of water, 
= 25 mol
We know that,




Hence, the vapour pressure of water is 17.44 mm of Hg.
31. 100 g of liquid A (molar mass 140 g) was dissolved in 1000 g of liquid B (molar mass 180 g ). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Ans. Number of moles of liquid A, 
= 0.714 mol
Number of moles of liquid B, 
= 5.556 mol
Then, mole fraction of A, 

= 0.114
And, mole fraction of B, 
= 0.886
Vapour pressure of pure liquid B, = 500 torr
Therefore, vapour pressure of liquid B in the solution,

= 
= 443 torr
Total vapour pressure of the solution, ptotal = 475 torr
Therefore, Vapour pressure of liquid A in the solution,

= 475 – 443
= 32 torr

Now, 

= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.
32. Vapor pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot p total’ p chloroform’ and p acetone as a function of x acetone. The experimental data observed for different compositions of mixture is.

011.823.436.050.858.264.572.1
acetone /mm Hg054.9110.1202.4322.7405.9454.1521.1
p chloroform/mm Hg632.8548.1469.4359.7257.7193.6161.2120.7

Ans. From the question, we have the following data

011.823.436.050.858.264.572.1
p acetone/mm Hg054.9110.1202.4322.7405.9454.1521.1
P chloroform/mm Hg632.8548.1469.4359.7257.7193.6161.2120.7
p tota(mm Hg)632.8603.0579.5562.1580.4599.5615.3641.8


It can be observed from the graph that the plot for the p total of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.
33. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of.
Ans. Molar mass of benzene 
= 
Molar mass of toluene 
= 
Now, no. of moles present in 80 g of benzene = 
= 1.026 mol
And, no. of moles present in 100 g of toluene = 
=1.087 mol
Therefore, Mole fraction of benzene, 
= 0.486
And, mole fraction of toluene, 
= 0.514
It is given that vapour pressure of pure benzene, 
And, vapour pressure of pure toluene, 
Therefore, partial vapour pressure of benzene,

= 24.645 mm Hg
And, partial vapour pressure of toluene, 

= 16.479 mm Hg
Hence, mole fraction of benzene in vapour phase is given by:


= 
= 0.599
= 0.6
34. The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 Kif the Henry’s law constants for oxygen and nitrogen are and respectively, calculate the composition of these gases in water.
Ans. Percentage of oxygen  in air = 20 %
Percentage of nitrogen in air = 79%
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, mm Hg = 7600 mm Hg
Therefore,
Partial pressure of oxygen, 
= 1520 mm Hg
Partial pressure of nitrogen, 
= 6004 mmHg
Now, according to Henry’s law:
For oxygen:


(Given )

For nitrogen:




Hence, the mole fractions of oxygen and nitrogen in water are  and  respectively.
35. Determine the amount of  (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at .
Ans. We know that,




V = 2.5 L
i = 2.47
T = (27 + 273)K = 300 K
Here,
R = 0.0821 L atm 

= 111g 
Therefore, 
= 3.42 g
Hence, the required amount of is 3.42 g.
36. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of in 2 liter of water at , assuming that it is completely dissociated.
Ans. When is dissolved in water,  and ions are produced.

Total number of ions produced = 3
Therefore, i =3
Given,
w = 25 mg = 0.025 g
V = 2 L
T = 250C = (25 + 273) K = 298 K
Also, we know that:
R = 0.0821 L atm 

Appling the following relation,