NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

Updated on Jul 9, 2020 - 1:32 p.m. IST by Sumit Saini

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics- Thermodynamics is the branch of science which deals with the transfer of energy or changes in energy in physical or chemical processes. This is an important chapter for class 11th students because some concepts of this chapter will also be applied in the next chapter which is equilibrium and also some in the 12th class chapters like chemical kinetics etc. In NCERT solutions for class 11 chemistry chapter 6 Thermodynamics, there are 22 questions in the exercise. The NCERT solutions for class 11 chemistry chapter 6 thermodynamics are prepared in a very comprehensive manner. These NCERT solutions of class 11 can help you in the preparation of class 11 final examination as well as in the competitive exams like JEE Mains, NEET, etc. If you are looking for an answer from any other chapter even from any other class then go with NCERT Solutions , there you will get all the answers of NCERT easily.

Thermodynamics enables us to study energy changes quantitatively and to make useful predictions. For these, we divide the universe into the system and the surroundings. In thermodynamics, the system is the part of the universe where observations are made and the remaining universe constitutes surroundings. After completing NCERT solutions for class 11 chemistry chapter 6 Thermodynamics you will be able to explain the terms like system, surroundings, and boundary; explain three types of system open, close and isolated systems; explain work, heat, and internal energy, etc. In this chapter, we will also discuss the first law of thermodynamics and second laws of thermodynamics and express them mathematically, internal energy, enthalpy, entropy, Gibbs energy. You will find NCERT solutions for class 11 chemistry chapter 6 Thermodynamics by scrolling down.

Important points of NCERT solutions for class 11 chemistry chapter 6 Thermodynamics-

1. System and Surroundings - in thermodynamics, the system is the part of the universe where observations are made and the remaining universe constitutes surroundings. The universe = The system+ The Surroundings .

2. The First Law of Thermodynamics- according to it under any circumstances, energy cannot be created and nor be destroyed, it can be converted from one form to another form with the interaction of work, heat and internal energy. Mathematically, this is represented as: $\Delta U=q+w$ , where w is the work done on or by the system, q is the heat exchanged between the system and its surroundings and ΔU is the total change in internal energy of the system .

3. The second law of thermodynamics- it states that entropy of the entire universe(total entropy of both the system and its surroundings), as an isolated system, will always increase over time.

Topics of NCERT Grade 11 Chemistry Chapter 6 Thermodynamics

6.1 Thermodynamic Terms

6.2 Applications

6.3 Measurement of âU and âH: Calorimetry

6.4 Enthalpy Change, âH of a Reaction – Reaction Enthalpy

6.5 Enthalpies for Different Types of Reactions

6.6 Spontaneity

6.7 Gibbs Energy Change and Equilibrium

NCERT solutions for class 11 chemistry chapter 6 Thermodynamics- Exercise Questions

Question 6.1 Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.

Answer :

A state function refers to the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of path.
Like- $p, V,T$ depends on the state of the system, not on the path.

So, the correct option is (ii)

Question 6.2 For the process to occur under adiabatic conditions, the correct condition is:

(i) $\Delta T = 0$
(ii) $\Delta p = 0$
(iii) q = 0
(iv) w = 0

Answer :

In adiabatic condition, the heat exchange between the system and the surrounding through its boundary is not allowed.
So, the correct option is (iii) $q$ = 0

Question 6.3 The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element

Answer :

The enthalpies of all elements in their standard states are Zero.
So, the correct option is (ii)

Question 6.4 $\Delta U ^\ominus$ of combustion of methane is $- X KJ mol ^{-1}$ . The value of $\Delta H ^\ominus$ is
(i) $= \Delta U ^\ominus$
(ii) $> \Delta U ^\ominus$
(iii) $< \Delta U ^\ominus$
(iv) = 0

Answer :

Equation of combustion of methane-

$CH_4(g)+\:2O_2(g)\rightarrow CO_2(g)\:+\:2H_2O(l)$

Since, $\Delta H^o=\Delta U^o+\Delta n_gRT$ and $\Delta U^o=-X\:KJ\:mol^{-1}$

from the equation $\Delta n_g=(n_p-n_R)=1-3=-2$

$\Delta H^o=-X-2RT$

$\Delta H^o<\Delta U^o$

so the correct option is (iii).

Question 6.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $-890.KJ mol ^{-1} , -393.5 KJ mol ^{-1} \: \:$ and $-285 .8 KJ mol ^{-1}$ respectively. Enthalpy of formation of $CH_4 (g)$ will be

(i) –74.8 $kJ mol^{-1}$

(ii) –52.27 $kJ mol^{-1}$

(iii) +74.8 $kJ mol^{-1}$

(iv) +52.26 $kJ mol^{-1}$

Answer :

$CH_4+2O_{2}\rightarrow CO_{2}+2H_{2}O\ \Delta H$ ................. -890.3 kJ/mol
$C+2O_{2}\rightarrow CO_{2}$ $\Delta H$ ..................-393.5 kJ/mol
$2H_{2}+O_{2}\rightarrow 2H_{2}O$ $\Delta H$ ..................-258.8 kJ/mol

So, the required equation is to get the formation of $CH_4 (g)$ by combining these three equations-

Thus, $C+2H_{2}\rightarrow CH_{4}$

$\Delta _fH_{CH_{4}} = \Delta _cH_c+2\Delta _cH_{H_2}$
= [-393.5 + 2(-285.8) + 890.3]
= -74.8 kJ/mol

therefore, the enthalpy of formation of methane is -74.8 kJ/mol. So, the correct option is (i)

Question 6.6 A reaction, A + B $\rightarrow$ C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(v) possible at any temperature

Answer :

For the reaction to be feasible the $\Delta G$ should be negative
$\Delta G = \Delta H - T\Delta S$

According to question,
$\Delta S$ = positive unit
$\Delta H$ = negative (as heat is evolved in the reaction)
Overall the $\Delta G$ is negative.

Therefore the reaction is possible at any temperature. So, the correct option is (iv)

Question 6.7 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Answer :

The first law of Thermodynamics state that,
$\Delta U = q+W$

where, $\Delta U$ = change in internal energy for the process
q = heat and W = work

Given that,
q = +701 J (heat is absorbed)
W = - 394 (work is done by the system)

Substituting the value in equation of first law we get,

$\Delta U = 701+(-394) =307J$

So, the change in internal energy for the process is 307 J

Question 6.8 The reaction of cyanamide, $NH_2 CN$ (s), with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be $-742.7 KJ mol ^{-1}$ at 298 K. Calculate enthalpy change for the reaction at 298 K.

$NH_2 CN ( g ) + \frac{3}{2} O_2 (g) \rightarrow N_2 + CO_2 ( g ) H_2 O ( l)$

Answer :

Given information,
$\Delta U = -742.7kJmol^{-1}$
T = 298 K
R = 8.314 $\times 10^{-3}$

$\Delta n_g =$ $n_g$ (products) $- n_g$ (reactants)
= (2 - 1.5)
= 0.5 moles

The enthalpy change for the reaction is expressed as;

$\Delta H = \Delta U+n_gRT$
where, $\Delta U$ = change in internal energy and
$\Delta n_g =$ change in no. of moles

By putting the values we get,
$\Delta H = (-742.7)+0.5(298)(8.314\times 10^{-3})$
$=-741.5 kJmol^{-1}$

Question 6.9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from $35 \degree C$ to $55 \degree C$ . Molar heat capacity of Al is $24 J mol ^{-1} K ^{-1}$

Answer :

We know that
$q = mc\Delta T$
m = mass of the substance
c = heat capacity
$\Delta T$ = change in temperature

By putting all these values we get,
$q = (\frac{60}{27}mol)(24 J/mol/K)(20K)$
= 1066.7 J
= 1.07 kJ

Question 6.10 Calculate the enthalpy change on freezing of 1.0 mol of water at $10.0 \degree C$ to ice at $- 10.0 \degree C$ .

$\Delta _{ fus }H = 6.03 KJ mol ^{-1}$ at $0 \degree C$ .

$Cp [H_2O(l)] = 75.3 J mol^{-1} K^{-1}\\\\.\: \: \: \: Cp [H_2O(s)] = 36.8 J mol^ { -1}K^{-1}$

Answer :

Total enthalpy change is equal to the summation of all the energy required at three different stages-
(i) from $10^o C$ to $0^o C$ of 1 mol of water (water $\rightarrow$ water )
(ii) from $0^o C$ to $0^o C$ of 1 mol of ice (water $\rightarrow$ ice)
(iii) from $0^o C$ to $-10^oC$ of 1 mole of ice (ice $\rightarrow$ ice)

So, the total enthalpy change
$\Delta H = C_p[H_2O(l)](\Delta T)+\Delta H_(freezing)+C_p[H_2O(s)]\Delta T$

$\\ = (75.3)(-10)+(-6.03\times 10^{3})+(36.8)(-10)\\ =-753-6030 - 368\\ =-7151 Jmol^{-1}\\ = 7.151kJ/mol$
Hence the total enthalpy change in the transformation process is 7.151 kJ/mol

Question 6.11 Enthalpy of combustion of carbon to $CO_2$ is $-393.5 KJ mol ^{-1}$ Calculate the heat released upon formation of 35.2 g of $CO_2$ from carbon and dioxygen gas.

Answer :

Formation of carbon dioxide from carbon and dioxygen reaction is-

$C+O_{2}\rightarrow CO_{2}$ ............. $\Delta H_f$ = -393.5 kJ/mol

1 mole of $CO_2$ = 44 g
So, the heat released in the formation of 44g of $CO_2$ = -393.5 kJ/mol
therefore, In 35.5 g of $CO_2$ the amount of heat released =
$\frac{-393.5}{44}\times 35.2 = -314.8 kJ/mol$

Question 6.12 Enthalpies of formation of $CO (g ) , CO_2 ( g ), N_ 2 O ( g ) \: \: and \: \: N_2 O_4 (g)$ are –110, – 393, 81 and 9.7 $KJ mol ^{-1}$ respectively. Find the value of $\Delta _ r H$ for the reaction:

$N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)$

Answer :

Given,
Enthalpies of formation of $CO (g ) , CO_2 ( g ), N_ 2 O ( g ) \: \: and \: \: N_2 O_4 (g)$ are –110, – 393, 81 and 9.7 $KJ mol ^{-1}$
$N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)$
We know that the $\Delta _rH=\sum \Delta _fH$ (product) $-\sum \Delta _fH$ (reactants)

For the above reaction,

$\Delta _rH =[{\Delta _fH(N_{2}O)+3\Delta _fH(CO_{2})}]-[\Delta _fH(N_{2}O_{4})+3\Delta _fH(CO)]$
Substituting the given values we get,
$\\= [81+3(-393)]-[9.7+3(-110)]\\ =-777.7 kJ/mol$
Thus the value of $\Delta _ r H$ of the reaction is -777.7 kJ/mol

Question 6.13 Given $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) ;\Delta rH = -92.4 kJ mol ^{-1}$ .What is the standard enthalpy of formation of $NH_3$ gas?

Answer :

the standard enthalpy of formation of any compound is the required change in enthalpy of formation of 1 mole of a substance in its standard form from its constituent elements.

Thus we can re-write the reaction as;
$\frac{1}{2}N_{2}+\frac{3}{2}H_{2}\rightarrow NH_{3}$

Therefore, standard enthalpy of formation of ammonia is = $\frac{1}{2}\Delta _rH^\Theta$
= 1/2 (-92.4)
= -46.2 kJ/mol

Question 6.14 Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

$CH_3OH (l) + 3/2 O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \Delta _r H = -726 kJ mol^{-1}\\\\ .\: \: \: \: C(graphite) + O_2(g) \rightarrow CO_2(g) ;\Delta_ cH = -393 kJ mol ^ {-1}\\\\ .\: \: \: H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l) ; \Delta_ f H = - 286 kJ mol^{-1}.$

Answer :

for the formation of $CH_3OH$ the reaction is ,

$C+2H_{2}O+\frac{1}{2}O_{2}\rightarrow CH_3OH$
This can be obtained by the following expressions-
required eq = eq (i) + 2 (eq iii) - eq (i)

$\Delta _fH[CH_{3}OH] = \Delta _cH^\Theta +2\Delta _fH[H_{2}O]-\Delta _rH^\Theta$
$=(-393) + 2(-286)-(-726)$
$=-239 kJmol^{-1}$

Question 6.15 Calculate the enthalpy change for the process $CCl_4(g)\rightarrow C(g) + 4 Cl(g)$ and calculate bond enthalpy of $C-Cl$ in $CCl_4(g)$
$\Delta _{vap} H ^{\ominus }(CCl_4) = 30.5 kJ mol^{-1}.\\\\ .\: \: \: \: \: \Delta _f H ^\ominus (CCl_4) = -135.5 kJ mol^{-1}.\\\\.\: \: \: \: \Delta _ a H ^ \ominus (C) = 715.0 kJ mol ^{-1} , where \: \: \Delta_ a H ^\ominus \: \: is\: \: enthalpy \: \: of \: \: atomisation\\\\.\: \: \: \: \Delta _ a H ^\ominus (Cl_2) = 242 kJ mol ^{-1}$

Answer :

We have the following chemical reactions equations-

$(1)\ CCl_{4}(l)\rightarrow CCl_{4}(g)$ ................ $\Delta _{vap}H^o = 30.5kJ/mol$
$(2)\ C(s)\rightarrow C(g)$ ........................ $\Delta _{a}H^o =715.0kJ/mol$
$(3)\ Cl2(g)\rightarrow 2Cl(g)$ .................. $\Delta _{a}H^o =242kJ/mol$
$(4)\ C(g)+4Cl\rightarrow CCl_4(g)$ ............ $\Delta _{f}H =-135.5kJ/mol$
The enthalpy change for the process $CCl_4(g)\rightarrow C(g) + 4 Cl(g)$ by the above reaction is calculated as;

$\Delta H =\Delta _aH^o(C)+2(\Delta _aH^o(Cl_{2}))-\Delta _{vap}H^o-\Delta _fH$
$= [(715)+2(242)-(30.5)-(-135.5)] kJ/mol$
$=1304 kJmol^{-1}$

And the bond enthalpy of C-Cl bond in $CCl_4$ (g) = 1304/4 = 326 kJ/mol

Question 6.16 For an isolated system, $\Delta U = 0$ , what will be $\Delta S$ ?

Answer :

Since $\Delta U = 0$
So, the $\Delta S>0$ (positive) therefore the reaction will be feasible.

Question 6.17 For the reaction at 298 K,
2A + B $\rightarrow$ C
$\Delta H = 400 kJ mol^ {-1 }\: \: and \: \: \Delta S = 0.2 kJ K ^{-1} mol^{-1}$
At what temperature will the reaction become spontaneous considering $\Delta H$ and $\Delta S$ to be constant over the temperature range.

Answer :

From the equation,

$\Delta G = \Delta H-T\Delta S$
Suppose the reaction is at equilibrium, So the change in temperature is given as;

$T=\frac{\Delta H -\Delta G}{\Delta S}$
$= \frac{400}{0.2} =2000K$ ( $\Delta G$ at equilibrium is zero)
To reaction should be spontaneous, $\Delta G$ should be neagtive. So, that for the given reaction T should be greater than 2000 K

Question 6.18 For the reaction, $2 Cl (g) \rightarrow Cl_2 ( g)$ , what are the signs of $\Delta H \: \: and \: \: \Delta S$ ?

Answer :

The given reaction represents the formation of a chlorine molecule from its atom. Bond formation taking place, therefore the energy is released during this. So, $\Delta H$ is negative.

Two moles of an atom have more randomness than the one mole of chlorine. So, spontaneity is decreased. Hence $\Delta S$ is negative.

Question 6.19 For the reaction
$2 A (g) + B ( g ) \rightarrow 2 D ( g)$
$\Delta U ^ \ominus = -10.5 KJ \: \: and \: \: \Delta S ^\ominus = - 44.1 JK^{-1}$
Calculate $\Delta G ^ \ominus$ for the reaction, and predict whether the reaction may occur spontaneously.

Answer :

Given reaction is $2 A (g) + B ( g ) \rightarrow 2 D ( g)$

We know that, $\Delta G^o = \Delta H^o-T\Delta S^o$
and $\Delta H^o = \Delta U^o+\Delta n_{g}RT$
here $\Delta n_{g}= 2 - (3) = -1$

and $\Delta U ^ \ominus = -10.5 KJ \: \: and \: \: \Delta S ^\ominus = - 44.1 JK^{-1}$

substituting the given values in equations-
$\Delta G = (-10.5+8.314\times 10^{-3}\times 298)-298(-44.1)$
$\\\Delta G = (-12.98 +13.14 )kJ\\ \Delta G= +0.16 kJ$

Hence the reaction will not occur spontaneously because the $\Delta G ^ \ominus$ value is positive for this reaction.

Question 6.20 The equilibrium constant for a reaction is 10. What will be the value of

$\Delta G ^ \ominus ? R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K$

Answer :

Given values are,
$R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K$
and equilibrium constant = 10

It is known that,
$\Delta G^o = -2.303\ RT \log K_{eq}$
$\\=-2.303(8.314)(300)\log 10\\ =-5744.14 Jmol^{-1}\\ =-5.744 kJmol^{-1}$

Hence the value of $\Delta G^o$ is -5.744 kJ/mol

Question 6.21 Comment on the thermodynamic stability of NO(g), given

$1/2 N_2(g) + 1/2 O_2(g)\rightarrow NO(g) ; \Delta _ rH ^ \ominus = 90 kJ mol ^{-1}\\\\ .\: \: \: NO(g) + 1/2 O_2(g) \rightarrow NO_ 2(g) : \Delta _ r H^ \ominus = -74 kJ mol ^{-1}$

Answer :

The formation of $NO$ is unstable because $\Delta _rH$ is positive it means heat is absorbed during the reaction. So, $NO$ (g) has higher energy than its reactants $N_{2}$ and $O_{2}$ .
On the other hand, $NO_{2}$ is stable because $\Delta _rH$ is negative means heat is released during the formation of $NO_{2}$ . It is stablised with minimum energy. Hence unstable $NO$ changes to stable $NO_{2}$ .

Question 6.22 Calculate the entropy change in surroundings when 1.00 mol of $H_2 O (l)$ is formed under standard conditions. $\Delta _f H ^ \ominus = -286 KJ mol ^ { -1}$

Answer :

-286 kJ/mol heat is evolved when 1 mol of the water molecule is formed. It means the same amount of energy is absorbed by the surrounding also.

$q_{surr}= +286$ kJ/mol
We know that,
Entropy changes for the surrounding is = q(surr)/ temp.
= 286000/ 298

= 959.73 J/mol/K

NCERT solutions for class 11 chemistry

Chapter 1	NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry
Chapter-2	NCERT solutions for class 11 chemistry chapter 2 Structure of Atom
Chapter-3	NCERT Solutions for class 11 chemistry chapter 3 Classification of Elements and Periodicity in Properties
Chapter-4	NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure
Chapter-5	NCERT solutions for class 11 chemistry chapter 5 States of Matter
Chapter-6	NCERT solutions for class 11 chemistry chapter 6 Thermodynamics
Chapter-7	NCERT solutions for class 11 chemistry chapter 7 Equilibrium
Chapter-8	NCERT solutions for class 11 chemistry chapter 8 Redox Reaction
Chapter-9	NCERT Solutions for class 11 chemistry chapter 9 Hydrogen
Chapter-10	NCERT solutions for class 11 chemistry chapter 10 The S-Block Elements
Chapter-11	NCERT solutions for class 11 chemistry chapter 11 The P-Block Elements
Chapter-12	NCERT Solutions for class 11 chemistry chapter 12 Organic chemistry- some basic principles and techniques
Chapter-13	NCERT solutions for class 11 chemistry chapter 13 Hydrocarbons
Chapter-14	NCERT solutions for class 11 chemistry chapter 14 Environmental Chemistry

NCERT solutions for class 11 subject wise

NCERT solutions for class 11 biology

NCERT Solutions for class 11 maths

NCERT solutions for class 11 chemistry

NCERT solutions for class 11 physics

Benefits of NCERT solutions for class 11 chemistry chapter 6 Thermodynamics

The comprehensive answers given in the CBSE NCERT solutions for class 11 chemistry chapter 6 Thermodynamics will be helpful in understanding chapter easily.
Revision will be quiet easier because with the detailed solutions will help you to remember the concepts and get very good marks in your class.
Homework problems won't bother you anymore, all you need to do is check the detailed NCERT solutions for class 11 chemistry Chapter 6 Thermodynamics and you are good to go.
If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

Question: Question: Whether the unit thermodynamics is helpful in higher studies?

Answer:

Yes, it is helpful to a large extent

Question: What is the official website of NCERT

Answer:

official website of NCERT: http://www.ncert.nic.in/

Question: What is the weightage of the chapter thermodynamics for CBSE board exam

Answer:

Weightage of thermodynamics for CBSE board exam is of 8 marks

Question: What is the weightage of NCERT Chapter Thermodynamics for NEET exam

Answer:

Weightage of Thermodynamics for NEET exam is 3%

Question: Where can I find complete solutions of NCERT class 11 Chemistry

Answer:

Refer to this link: https://school.careers360.com/ncert/ncert-solutions-class-11-chemistry

Question: What is the weightage of NCERT Chapter Thermodynamics for JEE mains

Answer:

8 marks

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NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

NCERT solutions for class 11 chemistry chapter 6 Thermodynamics- Exercise Questions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

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