Solutions Class 12 Notes JEE Advanced Chemistry [Free PDF Download]

Single Correct Type

1. A pressure cooker reduces cooking time for food because

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(A) Boiling point of water involved in cooking is increased

(B) the higher pressure inside the cooker crushes the food material

(C) cooking involved chemical changes helped by a rise in temperature

(D) heat is more evenly distributed in the cooking space

2. Which liquid pairs show a positive deviation from Raoult’s law?

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(A) Water-hydrochloric acid

(B) Acetone-chloroform

(C) Water-nitric acid

(D) Benzene-methanol

3. An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is :

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(A) 0.33

(B) 0.50

(C) 0.67

(D) 0.80

4. Haemoglobin and gold sol are examples of :

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(A) positively charged sols

(B) negatively charged sols

(C) positively and negatively charged sols, respectively.

(D) negatively and positively charged sols, respectively.

5. The size of a raw mango shrinks to a much smaller size when kept in a concentrated salt solution. Which one of the following processes can explain this?

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(A) Osmosis

(B) Reverse osmosis

(C) Diffusion

(D) Dialysis

6. Two open beakers, one containing a solvent and the other containing a mixture of that solvent with a non-volatile solute, are seated in a container. Over time :

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(A) The volume of the solution increases and the volume of the solvent decreases

(B) The volume of the solution does not change and the volume of the solvent decreases

(C) The volume of the solution and the solvent does not change

(D) The volume of the solution decreases and the volume of the solvent increases

7. Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?

(A) hydrazine

(B) glucose

(C) glycine

(D) ${\rm{KHS}}{{\rm{O}}_{\rm{4}}}$

8. For which of the following parameters the structural isomers C2H5OH and CH3OCH3 would be expected to have the same values? (Assume ideal behaviour)

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(A) Heat of evaporation

(B) Gaseous densities at the same temperature and pressure

(C) Boiling points

(D) Vapour pressure at the same temperature

9. Considering hydrogen bonds' formation, breaking and strength, predict which of the following mixtures will show a positive deviation from Raoult’s law?

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(A) Methanol and acetone.

(B) Chloroform and acetone.

(C) Nitric acid and water. 

(D) Phenol and aniline

10. Which of the following aqueous solutions should have the highest boiling point?

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(A) 1.0 M ${\rm{mol~kg}}{{\rm{K}}^{{\rm{ - 1}}}}$

(B) 1.0 M ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$

(C) 1.0 M ${\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{O}}_{\rm{3}}}$

(D) 1.0 M ${\rm{KN}}{{\rm{O}}_{\rm{3}}}$

11. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ____________.

(A) low temperature

(B) low atmospheric pressure

(C) high atmospheric pressure

(D) both low temperature and high atmospheric pressure

12. In comparison to $0.01{\rm{M}}$ a solution of glucose, the depression in freezing point of a $0.01{\rm{M}}{\rm{MgC}}{{\rm{l}}_{\rm{2}}}$ the solution is _____________.

(A)  The same

(B)  About twice

(C)  About three times

(D)  About six times

Multiple Correct Type

13. In the depression of freezing point experiment, it is found that the

(A)  Vapour pressure of the solution is less than that of pure solvent.

(B)  Vapour pressure of the solution is more than that of pure solvent

(C)  Only solute molecules solidify at the freezing point.

(D)  Only solvent molecules solidify at freezing.

14. Which of the following factor (s) affect the solubility of a gaseous solute in the fixed volume of liquid solvent?

1. Nature of solute

2. Temperature

3. Pressure 

(A) (a) and (c) at constant T

(B) (a) and (b) at constant P

(C) (b) and (c) only

(D) (c) only

15. When two benzene molecules are together, the intermolecular forces are almost as strong as when two toluene molecules are together. Which of the following statements about a benzene and toluene mixture is not true?

(A) ${\Delta _{{\rm{mix}}}}H = $ zero

(B) $\Delta _{{\rm{mix}}}^{}V = $ zero

(C) These will form a minimum boiling azeotrope.

(D) These will not form an ideal solution.

16. Mixture (s) showing positive deviation from Raoult’s law at 35oC is (are) 

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(A) Carbon tetrachloride + methanol

(B) Carbon disulfide + acetone

(C) Benzene + toulene

(D) Phenol + aniline

17.Van’t Hoff factor $\left( i \right)$ is given by the expression _____________.

(A) $i = \dfrac{{{\rm{ Normal~molar~ mass }}}}{{{\rm{ Abnormal ~molar~ mass }}}}$ 

(B) $i = \dfrac{{{\rm{ Abnormal~ molar~ mass }}}}{{{\rm{ Normal~ molar~mass }}}}$

(C) $i = \dfrac{{{\rm{ Observed~colligative~property }}}}{{{\rm{ Calculated~ colligative~ property }}}}$ 

(D) $i = \dfrac{{{\rm{ Calculated~colligative~property }}}}{{{\rm{ Observed~colligative~ property }}}}$

18. Isotonic solutions must have the same _____________.

(A) Solute

(B) Density

(C) Elevation in boiling point

(D) Depression in freezing point

19. Which of the following binary mixtures compositions in the liquid and vapour phases will be the same?

(A) Benzene - Toluene

(B) Water - Nitric acid

(C) Water - Ethanol

(D)  n-Hexane - n-Heptane

20. In isotonic solutions ________________.

(A) Solute and solvent both are same

(B) Osmotic pressure is same

(C) Solute and solvent may or may not be same

(D) Solute is always same solvent may be different

21. Which of the curves best describes the variation in total vapour pressure vs solution composition for a binary perfect liquid solution?

Total Vapour Pressure Versus Composition Of Solution

22. Colligative properties are observed when _____________. 

(A) A non-volatile solid is dissolved in a volatile liquid.

(B) A non-volatile liquid is dissolved in another volatile liquid.

(C) A gas is dissolved in non-volatile liquid.

(D) A volatile liquid is dissolved in another volatile liquid

Integer Type

23. A solution of 6% table sugar $\left( {{{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{22}}}}{{\rm{O}}_{{\rm{11}}}}} \right)$ is isotonic with 0.92% solution of X. Calculate the molar mass of substance X.

24. The vapor pressure for pure water is 23.76 torr at ${25^ \circ }{\rm{ C,}}$ and if a certain amount of NaCl is added to pure water, the vapor pressure of the solution becomes 19.6 torr. What will be the ratio of the values of the lowering in vapour pressure to the relative lowering in vapour pressure? Round off to the nearest integer.

25. The vapour pressure of pure benzene is 639.7 mm of mercury and the vapour of a solution of a solute in benzene at the same temperature is 631.9 mm of mercury. Calculate this molality of the solution.

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1. (A)

The temperature rises along with the increase in pressure. Thus, the water's boiling point increases in a pressure cooker as the pressure increases.

2. (D)

Solutions with solvent-solvent and solute-solute interactions that are greater than the solvent interactions exhibit positive deviations. The interactions between molecules grow weaker in such a solution. As a result, their tendency to escape increases, increasing their partial vapour pressures.

Intermolecular hydrogen bonds are present in benzene and methanol solutions.

In this mixture, benzene molecules sandwich ethanol molecules, weakening the interactions between the molecules. The vapour pressure rises as a result.

3. (B)

Assume that the degree of dissociation is $\alpha .$ 

MX2 $\longrightarrow$ M2+ + 2X-.

Thus, after dissociation total number of moles formed (n) = 3.

Now, the formula of degree of dissociation is $\begin{array}{c}\alpha  = \dfrac{{i - 1}}{{n - 1}}\\ = \dfrac{{2 - 1}}{{3 - 1}}\\ = 0.50\end{array}$

4. (C)

Haemoglobin is a positively charged sol because Heparin is the source of lack of coagulation. Gold sol is negatively charged sol.

5. (A)

Osmosis causes a net transfer of water molecules from the raw mango to the salt solution, causing the raw mango to shrivel in the solution.

6. (A)

The vapour pressure for solutions containing non-volatile solutes will decrease. Therefore, there will be a transfer of solvent molecules from pure solvent to solution, resulting in a decrease in the volume of the beaker containing pure solvent and an increase in the volume of the beaker containing solution.

7. (D)

The depression in freezing point can be expressed as, $\Delta {T_f} \propto i \propto m$

Here, i is the van’t Hoff factor and m is the molality.

Higher the value of i, higher will be the value of depression in freezing point.

Since the value of Van't Hoff factor is highest for ${\rm{KHS}}{{\rm{O}}_{\rm{4}}}$. Therefore, the colligative property will be highest for ${\rm{KHS}}{{\rm{O}}_{\rm{4}}}$.

8. (B)

At the same temperature and pressure, the gaseous densities of ethanol and dimethyl ether would be the same. Due to the H-bonding in ethanol, the heat of vaporisation, V.P., and b.pts will differ.

9. (A)

(A) Positive deviation from Raoult’s law refers to the observed vapour pressure is higher than the predicted vapour pressure and the solute-solvent interaction is weaker than the solute-solute and solvent-solvent interactions i.e.$A - B < A - A$, $A - B < B - B$

Pure methanol has hydrogen bonding. On addition of acetone, it breaks the hydrogen bonds between the acetone molecules and the interaction between the solute-solvent molecules is weaker than compared to solute-solute and solvent-solvent interactions. Therefore, it will show a positive deviation from Raoult’s Law.

(B) Positive deviation from Raoult’s law refers to the observed vapour pressure is lower than the predicted vapour pressure and the solute-solvent interaction is stronger than the solute-solute and solvent-solvent interactions i.e.$A - B > A - A$, $A - B > B - B$

Chloroform and acetone show negative deviations from Raoult’s law as solute-solvent interactions are stronger than solute-solute and solvent-solvent interactions. Therefore, this option is incorrect.

(C) Positive deviation from Raoult’s law refers to the observed vapour pressure is lower than the predicted vapour pressure and the solute-solvent interaction is stronger than the solute-solute and solvent-solvent interactions i.e.$A - B > A - A$, $A - B > B - B$.

Nitric acid and water show negative deviations from Raoult’s law as solute-solvent interactions are stronger than solute-solute and solvent-solvent interactions. Therefore, this option is incorrect.

(D) Positive deviation from Raoult’s law refers to the observed vapour pressure is lower than the predicted vapour pressure and the solute-solvent interaction is stronger than the solute-solute and solvent-solvent interactions i.e.$A - B > A - A$, $A - B > B - B$.

Phenol and water show negative deviations from Raoult’s law as solute-solvent interactions are stronger than solute-solute and solvent-solvent interactions. Therefore, this option is incorrect.

10. (B)

(A) Van’t Hoff factor denotes the number of ions of the compound formed when an ionic compound dissociates. It depends on the number of solute particles in the solution. Elevation in boiling point is one of them and it depends on the Van’t Hoff factor ‘i’ which depends on the number of solute particles. So we can say more is the number of ions in solutions, more is the boiling point of the solution.

$\Delta {T_b} = i{K_b}m$

$\Delta {T_b}$ is the elevation in boiling point

$i$ is the Van’t Hoff factor

${K_b}$ is the molal elevation constant

$m$ is the molality

Since molality and molal elevation constant ${K_b}$ is constant for all, elevation in boiling point depends on the Van’t Hoff factor $i$, $n = 2$.

The number of ions is two. Therefore, the boiling point is less so the option is incorrect.

(B) The number of ions is three so it will have the highest boiling point. Therefore, this option is correct.

(C) The number of ions here is two so it has less boiling point. Therefore, this option is incorrect.

(D) The number of ions here is two so it has less boiling point. Therefore, this option is incorrect.

11. (B)

At higher altitudes the atmospheric pressure is low but the body's temperature is the same. The oxygen is less at higher altitudes and the decreased atmospheric pressure is the reason for the release of oxygen from blood therefore the concentration of oxygen in blood and tissues of people is less.

12. (C)

The formula gives the depression in freezing point, $\Delta {T_f} = i{K_f} \times $ molality, where i is the dissociation constant or Van’t hoff factor.

Compared to solution of glucose whose value of $i$ is 1, the Van’t hoff factor for ${\rm{MgC}}{{\rm{l}}_{\rm{2}}}$ is 3. So the depression in freezing point will be three times that of glucose.

13. (A) and (D)

The vapour pressure of a solution is lower than that of a pure solvent in the depression in freezing point experiment. At freezing point, only solvent molecules solidify.

The point on a graph where the liquid's vapour pressure curve crosses the solid's vapour pressure curve is the temperature when the liquid's vapour pressure equals the solid's vapour pressure. A nonvolatile solute's freezing point is lowered when it is dissolved in a liquid.

14. (A) and (B)

The solubility of a gaseous solute in the fixed volume of liquid solvent always depends upon the nature of the solute. The solubility of the solute in fixed volume also depends on the pressure at constant temperature and temperature at constant pressure.

15. (C) and (D)

According to Raoult's law, if the intermolecular forces between two benzene molecules are the same as two toluene molecules, then in a mixture of benzene and toluene, the intermolecular forces between benzene and toluene will also be nearly equal. The solution will be an ideal solution as it obeys Raoult’s law.

So, for an ideal solution, ${\Delta _{{\rm{mix}}}}H = $ zero and $\Delta _{{\mathop{\rm mix}\nolimits} }^{}V = $ zero will be true. Hence, as this solution is ideal, the minimum boiling azeotropes will not be formed.

16. (A) and (B)

It is an optimum solution when the intermolecular attraction between two components A and B in a combination is the same as the attraction between A and A or B and B. The mixture is more vaporised and the bp is dropped when the intermolecular attraction between A and B in a mixture is smaller than that between A and A or B and B. Raoult's law is being positively deviated from in this instance. The mixture is less vaporised and the bp is higher when the intermolecular attraction between A and B is greater than that between A and A or B and B. A case of negative deviation exists here.

(A) Hydrogen bonds connect the molecules of methanol ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}{\rm{.}}$ The extent of H-bonding decreases when ${\rm{CC}}{{\rm{l}}_{\rm{4}}}$ and ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}$ are combined. More vaporisation of the mixture results in a positive divergence from Raoult's law.

(B) Because of the dipole-dipole interaction, acetone molecules are more attracted to one another. Positive deviation is caused by a decreased interaction caused by ${\rm{C}}{{\rm{S}}_{\rm{2}}}{\rm{.}}$

(C) Benzene and toluene combined to make the optimum solution.

(D) Intermolecular H-bonding increases the contact between phenol and aniline. Negative deviation follows.

17. (A) and (C)

Van’t Hoff factor is the measure of the association or dissociation of solute particles. 

The expressions for Van’t Hoff factor are:

$i = \dfrac{{{\rm{ Normal~molar~mass }}}}{{{\rm{ Abnormal~molar~mass }}}}$

And, $i = \dfrac{{{\rm{ Observed~ colligative~ property }}}}{{{\rm{ Calculated~colligative~ property }}}}$

18. (C) and (D)

Isotonic solutions are defined as those solutions which have the same osmotic pressure and same concentration. The two colligative properties, elevation in boiling point and depression in freezing point depend on the concentration.

For an isotonic solution the molar concentration is same, so the elevation in boiling point and depression in freezing point also must be same.

19. (B) and (C)

The mixture having the same liquid composition as the vapour phase is known as azeotropes. The property of azeotropic mixture is that it boils are the same temperature.

The solutions, water-nitric acid and water-ethanol are non-ideal solutions which have the same composition in liquid and vapour phase, hence they are azeotropes. While benzene-toluene and n-hexane-n-heptane are ideal solutions, they will not behave as azeotropes.

20. (B) and (C)

Isotonic solutions are defined as those solutions having the same osmotic pressure. Although the osmotic pressure of the two solutions is same but the solute and the solvents particles may or may not be same.

21. (A) and (D)

For a binary ideal liquid solution, the variation in total vapour pressure versus composition of solution is shown by the following graphs:

According to the above graphs, the total vapour pressure shows a linear composition of the solution. As the mole fraction of more volatile component increases, the total vapour pressure also increases. 

22. (A) and (B)

Colligative properties are observed in two conditions, first, when a non- volatile solid is dissolved in a volatile liquid and second, when a non-volatile liquid is dissolved in another volatile liquid. In both the cases, the solution deviates from the ideal behaviour and its vapour pressure changes, which changes the colligative properties. 

23. 6% table sugar solution means 6 g of table sugar is present in 100 ml or 0.1 L of the solution.

0.92% solution of X means 0.92 g of X is present in 100 ml or 0.1 L of the solution.

The number of moles of table sugar $\left( {{n_1}} \right)$ is calculated as $\begin{array}{*{20}{c}}{{n_1}}& = &{\dfrac{{{\rm{Given mass of table sugar}}}}{{{\rm{Molar~ mass~ of ~table~ sugar}}}}}\\{}& = &{\dfrac{{{\rm{6 g}}}}{{34{\rm{2 g mo}}{{\rm{l}}^{ - 1}}}}{\rm{                             }}}\end{array}$.

The number of moles of X $\left( {{n_2}} \right)$ is calculated as $\begin{array}{*{20}{c}}{{n_2}}& = &{\dfrac{{{\rm{Given~ mass~ of ~X}}}}{{{\rm{Molar ~mass ~of ~X}}}}}\\{}& = &{\dfrac{{{\rm{0}}{\rm{.92 g}}}}{{{\rm{Molar~ mass~ of ~X}}}}}\end{array}$.

The molarity of a solution is defined as the number of moles of solute dissolved per litre of solution.

The concentration of table sugar $\left( {{C_1}} \right)$ is calculated as $\begin{array}{*{20}{c}}{{C_1}}& = &{\dfrac{{{\rm{Moles~  of ~ table sugar}}}}{{{\rm{Volume~ of ~solution}}\left( {{\rm{in L}}} \right)}}}\\{}& = &{\dfrac{{\dfrac{{{\rm{6 g}}}}{{{\rm{342 g mo}}{{\rm{l}}^{ - {\rm{1}}}}}}}}{{{\rm{0}}{\rm{.1 L}}}}{\rm{           }}}\end{array}$.

Similarly, the concentration of X $\left( {{C_2}} \right)$is calculated as $\begin{array}{*{20}{c}}{{C_2}}& = &{\dfrac{{{\rm{Moles ~ of~  X}}}}{{{\rm{Volume~ of ~solution}}\left( {{\rm{in L}}} \right)}}{\rm{                }}}\\{}& = &{\dfrac{{\dfrac{{{\rm{0}}{\rm{.92 g}}}}{{{\rm{Molar~ mass~ of ~X}}}}}}{{{\rm{0}}{\rm{.1 L}}}}{\rm{           }}}\end{array}$.

Isotonic solutions have the same osmotic pressure $\left( \Pi  \right).$

So, the osmotic pressure of table sugar solution is equal to that of X solution. Thus, $\begin{array}{*{20}{c}}{{C_1}RT}& = &{{C_2}RT}\\{\dfrac{{\dfrac{{{\rm{6 g}}}}{{3{\rm{42 g mo}}{{\rm{l}}^{ - 1}}}}}}{{{\rm{0}}{\rm{.1 L}}}}}& = &{\dfrac{{\dfrac{{{\rm{0}}{\rm{.92 g}}}}{{{\rm{Molar~mass~ of ~X}}}}}}{{{\rm{0}}{\rm{.1 L}}}}}\\{\dfrac{{{\rm{5 g}}}}{{3{\rm{42 g mo}}{{\rm{l}}^{ - 1}}}}}& = &{\dfrac{{{\rm{0}}{\rm{.92 g}}}}{{{\rm{Molar~ mass~ of~ X}}}}}\end{array}$

Solving further,

$\begin{array}{*{20}{c}}{{\rm{Molar ~mass~ of~ X}}}& = &{\dfrac{{\left( {0.92{\rm{ g}}} \right) \times \left( {34{\rm{2 g  mo}}{{\rm{l}}^{ - 1}}} \right)}}{{{\rm{6 g}}}}}\\{}& = &{52.44{\rm{ g  mo}}{{\rm{l}}^{ - 1}}{\rm{                 }}}\end{array}$

Therefore, the molar mass of substance X is

$52.44{\rm{ g  mo}}{{\rm{l}}^{ - 1}}.$

24. The lowering of vapour pressure can be expressed as

${\rm{lowering ~in ~vapour~ pressure}}\left( {\Delta P} \right) = {P_{{\rm{solvent}}}} - {P_{{\rm{solution}}}}$

Here, ${P_{{\rm{solvent}}}}$ is the vapour pressure of pure solvent and ${P_{{\rm{solution}}}}$ is the vapour pressure of the solution.

The lowering in vapour pressure after adding NaCl is

$\begin{array}{*{20}{c}}{{\rm{lowering~ in~ vapour~ pressure}}}& = &{\left( {23.76 - 19.6} \right){\rm{ torr}}}\\{}& = &{4.16{\rm{ torr}}}\end{array}$

Again, ${\rm{relative ~lowering~ in ~vapour ~pressure}} = \dfrac{{{P_{{\rm{solvent}}}} - {P_{{\rm{solution}}}}}}{{{P_{{\rm{solvent}}}}}}$

Putting the values,

$\begin{array}{*{20}{c}}{{\rm{relative~ lowering~ in~ vapour~ pressure}}}& = &{\dfrac{{23.76 - 19.6}}{{23.76}}}\\{}& = &{0.17}\end{array}$

The ratio of the lowering in vapor pressure to the relative lowering in vapour pressure is

$\begin{array}{*{20}{c}}{\dfrac{{{\rm{lowering~ in ~vapour~ pressure}}}}{{{\rm{relative~ lowering~ in ~vapour~ pressure}}}}}& = &{\dfrac{{4.16}}{{0.17}}}\\{}& \approx &{\dfrac{{25}}{1}}\end{array}$

Therefore, the ratio of the values of the lowering in vapour pressure to the relative lowering in vapour pressure is 25:1.

25. It is given that the vapour pressure of pure benzene is ${p^0} = 639.7{\rm{ mm Hg}}$ and that of solution is $p = 631.9{\rm{ mm Hg}}{\rm{.}}$

If ${n_1}{\rm{ \ and \ }}{n_2}$ be the number of moles of benzene and solute respectively, then according to Raoults' law, $\dfrac{{{p^0} - p}}{{{p^0}}} = {x_2} \approx \dfrac{{{n_2}}}{{{n_1}}}$

Substitute the values,

$\begin{array}{c}\dfrac{{639.7 - 631.9}}{{639.7}} = \dfrac{{{n_2}}}{{{n_1}}}\\{n_2} = 0.0122 \times {n_1}\end{array}$

The amount of benzene in solution is calculated as ${n_1}{\rm{mol}} = 78 \times {n_1}{\rm{g}}$

The molality of the solution is calculated as $\begin{array}{c}m = \dfrac{{{n_2} \times 1000}}{{78 \times {n_1}}}\\ = \dfrac{{0.0122 \times {n_1} \times 1000}}{{78 \times {n_1}}}\\ = 0.156{\rm{ mol}}/{\rm{kg}}\end{array}$

Importance of JEE Advanced Class 12 Chemistry Solutions Important Questions

Solutions JEE Advanced questions have been designed to enrich the knowledge of students and test their understanding of the chapter. The more they solve these important questions, the more chances they have of clearing the JEE Advanced examination. These important questions from the chapter are based on the topics that have been discussed in Solutions.

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With the help of JEE Advanced Chemistry Chapter 2 Solutions important questions, students will be able to solve complicated numerical questions and score great marks in the examination. With a better grasp of the concept, students will be able to understand what questions might appear in the exam and prepare accordingly.

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