CHSE_CBSE_ 2_3_ Marks Chemistry Questions and Answers (TOPIC: Solution)

 2 or 3 marks CHSE/CBSE Chemistry Questions and Solutions 

(Topic: Solution) 

1. State and explain  Henry's law.

Ans: The dependence of solubility of a gas in a solvent on pressure can be quantitatively expressed by Henry’s Law. This law can be expressed in a number of ways:

(a) The solubility of a gas (in mol ltr^-1) in a liquid at a particular temperature is directly proportional the partial pressure of the gas in equilibrium with the liquid at that temperature.

     Mathematically, s α p   => s = k p, where s= solubility, Unit of  k = mol L^-1 atm^-1

(b) The amount of a gas dissolved (expressed in terms of mole fraction) in a definite volume of solvent at constant temperature is directly proportional to the partial pressure of the gas present in equilibrium with the solvent at that temperature.

    Mathematically, p_g = k_H X_g

    Where, KH is the Henry law constant. A gas having high solubity comparision to other has lower KH value.

2. What is relative lowering of vapour pressure?

Ans: The lowering in vapour pressure of a solution w.r.t. the vapour pressure of the pure solvent is called the rlative lowering in vapour pressure.

    The difference between the vapour pressures of a pure solvent and the corresponding solution is called the lowering in vapour pressure. When this lowering in vapour pressure is compared to the vapour prssure of the pure solvent, it is called teh relative  loweriing in vapour pressure.

    Mathematically, (p⁰_A - p^s) / p⁰ = X_solute. Where p⁰_A - p^s  is the lowering in vapor pressure, and thus (p⁰_A - p^s) / p⁰ is called the relative lowering in vapor pressure which is equal to the mole fraction of the solute.

3. What is the reason of elevation in boiling point?

Ans: When a nonvolatile solute is added to a pure solvent, the vapor pressure of solution thus formed becomes less than that of the pure solvent. Now the solution should be heated more in comparison to the pure solvent in order to get enough vapors whose pressure becomes equal to the atmospheric pressure in order to attain the boiling point. Therefore a solution always boils at a higher temperature than the pure solvent.

    The increase in the boiling temperature of a solution w.r.t. to the boiling point of the corresponding pure solvent is called the elevation in boiling point.

4. State Roult's law for solution containing volatile solute.

Ans: In Raoult’s law, the relationship between the vapor pressure of the solution and the corresponding pure solvent and the amount of solute dissolved in the solvent has been established. This can be discussed under two subsections:

a.        Solution containing volatile solute: The vapor pressure of a component in a volatile mixture is directly proportional to the mole fraction of that component.

      Mathematically,  p_A  α  X_A   and p_B  α  X_B  =>    p_A  = p⁰_A  X_A   and p_B  =  p⁰_B  X_{B ,} where p_A and p_B  are the vapor pressures due to component A and B in the volatile mixture. X_A and X_b are the mole fractions of the component A and B. p⁰_A and p⁰_B are the vapor pressure of pure solvent A and B.

b.        Solution containing nonvolatile solute: The vapor pressure of a solution is directly proportional to the mole fraction of the solvent.

      Mathematically, p^s α X_A     =>  p^s = p⁰_A X_A   , where p^s = v.p of the solution and X_A is the mole fractions of the solvent. p⁰_A is the vapor pressure of pure solvent. 

5. Draw the Graph of elevation in boiling point.

Ans: 

Where, P0 = V.P of pure solvent, T0 = B.P of pure solvent and T1 and T2 are the boling temperatures of solution 1 and solution 2. 

T1 - T0 = Elevation in boling point of solution 1

T2 - T0 = Elevation in boling point of solution 2

6. Draw the graph of depression in freezing point.

Ans:

Where, P0 = V.P of pure solvent, T0 = F.P of pure solvent and T1 and T2 are the freezing temperatures of solution 1 and solution 2.

T0 - T1 = Depression in Freezing point of solution 1

T0 - T2 = Depression in Freezing point of solution 2

7. What is molal elevation constant?

Ans: Molal elevation constant or Ebbulioscopic constant is defined as the elevation in boiling point of one molal solution. 

We know, ΔT_b  = k_b m’ . Thus when  molality = 1 m, ΔT_b  = k_{b .}  

Accordingly we wrote the definition in that way.

The unit of  molal elevation coinstant is kelvin kg per mole and this constant for a particular solvent. The molal elevation constant of water is 0.52 kelvin kg per mole.

8. What is molal depression constant?

Ans: Write similarly

9. State Avogadro vant Hoff law of osmotic pressure.

Ans: Equal volume of all solutions having the same osmotic pressure and temperature contain the same number of solute molecules or particles.

     For example, 1 litre solution of 1 M urea and 1 litre of M/2 NaCl solution exert same osmotic pressure at same temperature, as they contain equal number of particles.

     Solutions having same osmotic pressure at same temperature are called isotonic or iso-osmotic solution.

10. How can you determine the molecular weight of an unknown solute from osmotic pressure of a solution.

Ans: By developing the van't Hoff's ideal equation for dilute solution further we can have a formula from which we can dtermine the molecular mass of an unknown solute:

πV = nRT, this is the van’t Hoff equation for dilute solution.

Again n, the number of moles = w/m (given mass of solute/molecular mass of solute)

=>   πV = (w/m)RT, knowing all other values we can calculate the molecular mass of an unknown solute using this formula.

11. What is the relationship between vapour pressure s of volatile component and molefeaction of them in their solution?

Ans: 

Roult's law for solution containing volatile solution is:     The vapor pressure of a component in a volatile mixture is directly proportional to the mole fraction of that component.

Mathematically,  p_A  α  X_A   and p_B  α  X_B  =>    p_A  = p⁰_A  X_A   and p_B  =  p⁰_B  X_{B ,} where p_A and p_B  are the vapor pressures due to component A and B in the volatile mixture. X_A and X_b are the mole fractions of the component A and B. p⁰_A and p⁰_B are the vapor pressure of pure solvent A and B.

      The vapour pressure of the solution will be equal to,

p^s  =   p_A + p_B   =   p⁰_A  X_A + p⁰_B  X_B  =  p⁰_A  X_A + p⁰_B  (1 – X_A)  = (p⁰_{A ­}- p⁰_B) X_A  + p⁰_B  

12. State Boyles vant Hoff law.

Ans: Write the statement and the mathematical formula.

13. State Charle's vant Hoff law.

Ans: Write the statement and the mathematical formula.

14. What is binary solution and what are its type?

Ans: A solution containing two components is called a binary solution. 

Depending on the physical states of the solvents, binary solutions can be classified into the following types:

1. Liquid Solution: Here the solvent is a liquid and solute can be either solid or liquid or gas. For example: Sugar in water, ethanol in water and oxygen in water.

2. Gaseous Solution: Here the solvent is a gas and solute can be either solid or liquid or gas. For Example: Camphor or chloroform or oxgen in nitrogen.

3. Solid Solution: Here the solvent is a solid and solute can be either solid or liquid or gas. For Example: Copper in gold, mercury in sodium (sodium amalgam) and hydrogen in paladium.

15. What is solid solution and what are its type?

Ans: 

Solid Solutions: The solution in which both the solute and solvent are solids is called a solid solution. There are two types of solid solutions:

(1)  In Substitutional Solid Solution, the solute particles occupy (take the place of) the position of solvent particles in a crystal lattice. In this case the atomic sizes of the solute and solvent particles are equivalent.  Example: Brass, Bronze, Alloys of Copper and nickel, Silver and gold, Molybdenum and tungsten, etc.

(2)  In Interstitial Solid Solution, the solute particles occupy (take the place of) the interstitial positions in the crystal lattice. These solutions are formed because the solute atoms are small enough to fit into the interstitial positions between the solvent atoms. Example: Carbon atoms occupying the interstitial positions among iron solvents atoms in steel, hydrogen atoms occupying the interstitial positions among palladium or platinum solvent atoms, etc.

16. 3g urea is dissolved in hundred ml water. find the freezing point of solution (kf = 1.86k kg per mole)

Ans: 

Since density of water = 1g/ml, mass of solvent = W =100g, weight of solute = w = 3g, freezing point of pur solvent = 0 deg centigrade = 273k, m = molecular mass of solute (urea) = 60,

ΔTf = (kf . w . 1000) / (m . W), Thus calculate ΔTf and subtract it from  273k to get the answer

17. What are the conditions to behave like Ideal  solutions. Give examples.

(This question carries in actual 3 or 5 marks)

Ans: The solutions which obey Raoult’s law and satisfy the following conditions are called ideal solutions.

Condition 1: Change in volume before and after mixing the solvents (to prepare the solution) is equal to zero.  i.e., ΔV_mix = 0

Conditions 2: Change in enthalpy before and after mixing the solvents (to prepare the solution) is equal to zero.  i.e., ΔH_mix = 0.

From Raoult's Law, p_A  = p⁰_A  X_A   and p_B  =  p⁰_B  X_{B ,} where p_A and p_B  are the vapor pressures due to component A and B in the volatile mixture. X_A and X_B are the mole fractions of the component A and B. p⁰_A and p⁰_B are the vapor pressure of pure solvent A and B.

Thus, P(total) = P_A+P_B  = P⁰_A  X_A   +  P⁰_B  X_B  it’s called an ideal solution (graph:1 below). This happens when force of attraction between component A and B (F_A-B) is equal to the force of attraction between component A and A (F_A-A) and the force of attraction between component B and B (F_B-B). i.e., F_A-B = F_A-A = F_B-B. Clearly very dilute solutions prepared from liquids having similar nature and structures are ideal. Example: 1. Mixture of methanol and ethanol 2. Mixture of n-hexane and n-heptane 3. Mixture of benzene and toluene.

18. What is osmosis and what are its type?

Ans: The movement of solvent molecules from less concentrated solution to the higher one when both the solutions are separated by a semepermeable membrane is called the osmosis.

          When solvent molecules pass out of the cell due to osmosis, it is called exo-osmosis and when passe into the cell, it is known as endo-osmosis. A peeled egg shrinks when placed inside concentrated NaCl solution. This is becase water content from inside the egg comes out off egg. This is the example of exoosmosis. a peeled egg swells when placed in pure water because water enters into the egg due to concentration difference. this is called endoosmosis.

19. What is Freezing point? Draw a graph for the depression in freezing point of a 

solution.

Ans: The temperature at which both solid and liquid co exist and both have the same 

vapour pressure is called the freezing point. 

Here, T0 = freezinf point of pure solvent,

T1 = Freezing point point of solution 1

T2 =  Freezing point point of solution 2

20. What is elevation in boiling point? Draw a graph for the elevation in boiling 

point of a solution.

Ans: Give definition and draw the graph.

21. Draw the graph for ideal solution.

22. Draw the graph for non ideal solution showing positive and negative deviation.

Ans:

Here ΔV_mix > 0 , ΔHmix  > 0 and P1 + P2 > Ptotal,  where ΔV_mix, ΔHmix, P1, P2 and Ptotal are change in volume after mixing, change in enthalppy afer mixing, v.p of liquid 1, v.p of liquid 2 and the total vapour pressure respectively.

23. What is azeotrope? Give example.

Ans: Ans: A mixture of two liquids which has a constant boiling point and composition throughout the distillation is called the Azeotrope. In other words, mixtue of liquids which boils at constant temperature and have same composition both in solid and liquid phase is called an azeotrope.

    If the mixture shows positive deviation from Raoult's law then the azetrope will boil at a temperature which is less than the boiling point of any of the component. This type of azeotrope is called positive azeotrope or minimum boiling azeotrope.

    On the other hand if the mixture shows negative deviation from Raoult's law then the azetrope will boil at a temperature which is higher than the boiling point of any of the component. This type of azeotrope is called negative azeotrope or maximum boiling azeotrope.

Example: The ethanol water mixture (95.63% C2H5OH + 4.37 % by mass H2O) is a minimum boiling azeotrope as it boils at 78.2 degree centigrade which is less than the boiling point of ethanol (78.4 unit) and water (100 unit)

24. Explain why freezing point depression of 0.1 M NaCl is nearly twice that of 0.1 M glucose solution.

Ans: Glucose is a nonelectrolyte where as NaCl is strong electrolyte which dissociates completely. The number of particles formed in 0.1M NaCl is almost double of that formed in 0.1M glucose solution. Since colligative properties are directly proportional to the number of particles. Thus double number of particle in 0.1M NaCl is the reason why its F.P is double of that glucose. 

25. What is osmotic pressure? Why is it a colligative property?

Ans: It can be defined as the excess pressure which must be exerted on the solution side  (more concentrated) in order to check the flow of solvent molecules towards the solution side (more concentrated) when two solutions are separated by a semi permeable membrane.

According to B.V law,   π α C, at constant temperature

According to C.V law,   π α T, at constant concentration

On combination of the above two laws,

π α CT    => π = CRT , where π, C, R and T are osmotic pressure, concentration, universal gas constant and temperature respectively.

Since concentration = n/v  => π = (n/V) RT  =>  πV = nRT, where n and v are number of moles and volume respectively. 

Again n = w/m (given mass of solute/molecular mass of solute)

=>   πV = (w/m)RT, 

Since osmotic pressure depends on the molecular mass of the solute, its a colligative property.

26. What weight of the non volatile urea must be dissolved in hundred gram of water in order to decrease its vapour pressure by 25%?

Ans: Let V.P of pure solvent = 100, Thus V.P of Solution = 100 - 25 = 75%

As we know, relative lowering in vapour pressure = (P0 - Ps) / P0 = wM / Wm

W = 100 gram, m = molecular mass of urea = 60, M = 18 and w = ?

Thus (100 - 75) / 100 = w . 18 / 100 . 60

Therefore, w =weight of urea = 83.3 g  

 27. What is van't Hoff factor? What is its value for the dissociation and association of the solute in solution?

Ans: The values of colligative properties are directly proportional to the number of particles in the solution. The actual number of particles in the solution change when association or dissociation of solute particles takes place. Accordingly, the molar mass of solute calculated in this way deviate from true molar mass hence called abnormal molar mass. The  van't Hoff factor ' i '  relates to the abnormal colligative properties hence abnormal molar mass.

    It is defined as the ratio of number of particles after association or dissociation to the actual number of solute particles taken.

    i = number of particles after association or dissociation / actual number of solute particles taken.

    i = Experimental colligative property / calculated colligative property

    i = calculated molecular mass / experimental molecular mass

The number of particles after association is less than the actual particles taken hence i < 1 

similarly in case of dissociation i > 1.

28. The osmotic pressure of a solution is 2.4 atm at 27 degree centigrade if its strength is 18 gram per litre find molecular mass of the solute.

Ans: π = 2.4 atm, T = 300k, w = 18 g, v = 1 litre, m = ?

π = wrt / mv, now put the values and get the answer.

29. Calculate the molality of KCl solution in water if its freezing point is 271 Kelvin.

Ans: T0 = 273k, T' = 271k, ΔTf = 273 - 271 = 2K, KCl gives 2 particles on dissociation, 

so i = 2, Kf = 1.86

ΔTf  = i Kf m'  ....... get the answer

No. 30 to 35 is for you. Solve yourself.

30. How many grams of glucose must be dissolved in 0.5 litre of water for its osmotic pressure to be the same as that of a solution of 12 gram urea in 250 ml water.

31. If mole fraction of a solvent is 0.25 and its vapour pressure in pure state is 150 mm find the vapour pressure of the solution.

32. 10 gram of non volatile solute was dissolved in hundred gram water at 20 degree centigrade. The vapour pressure was lowered from 17.35 to 17.2 3mm calculate molecular mass of the solute.

 

33. A 5% solution of sucrose is isotonic with 3% aq. solution of another substance A. Find molecular mass of A. 

34. Dissolution of 10 gram solute in hundred gram of benzene increases its boiling point by 1 degree centigrade calculate molecular mass of solute (kb = 2.35 Kelvin kg per mole)

35. How much urea must be added to 500 gram water so that it will boil at 374 Kelvin. (Kb = 0.52 Kelvin kg per mole)