Bihar Board - Class 12 Chemistry -Chapter 2: Solutions Long Answer Question

                                   Long Question Answer 

Q. 1. Define the following terms :
(i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.

Ans⇒ (i) Mole fraction : The ratio of the moles of a component A to be the total moles of the solution is called the mole fraction of A . It is denoted by symbol x . For a binary solution of components A and B .
Mole fraction of A= XA .

= Moles of ATotal moles of solution = nAnA + nB

Mole fraction of B =XB

= Moles of BTotal moles of solution = nBnA + nB

(ii) Molality : The moles of the solute dissolved in one kilogram of the solvent is called the molality of the solution.

Molarity= m = Moles of BMass of solvent in kg

Moles =Moles of BMolar Mass of B = WBMB

m = WBMB WA

where WA = Mass of solvent in kg.

(iii) Molarity : The concentration expressed as the moles of solute per litre of solution is the molarity. It is denoted by M .

Molarity = Mole of solutionLitres of solution = nBV

A solution of one molar concentration is represented by 1 M, a two molar solution by 2 M and a one tenth molar solar solution is represented by 0.1 M .

(iv) Mass percentage : It is the amount of solute in gram’s dissolved per 100 g solution. E.g. 5 %  solution sodium carbonate mass 5 g of solid sodium carbonate is present is 100 g solution.

Mass % = Moles of soluteMass of solution 100

Q. 2. Which type of deviation is shown by the solution formed by mixing cyclohexane and ethanol ?

Ans⇒ Hydrogen bonds exist among the molecules of ethanol (i.e., strong forces). When cyclohexane is added to ethanol, the cyclohexane molecule will come in between the ethanol molecules and disturb the hydrogen bondings. The forces between ethanol and cyclohexane will be weaker than the previous forces (H-bonding), hence there will be positive deviation.

Q. 3. (a) What is reverse osmosis ?

(b) What is the boiling point of glucose solution containing 18 gm of glucose in 100 gm of H2O ? Molar elevation constant of H2O  is 0.52 k kg mol-1 .

Ans⇒ (a) Reverse osmosis : If external pressure greater than osmotic pressure is applied on solution, the flow of solvent molecules can be made to proceed from the solution towards pure solvent. This type of osmosis is termed as reverse osmosis. Reverse osmosis is used for the distillation of seawater for getting fresh drinking water.

(b) We know that

Tb = Kb WB 1000MB WA

MBWA Molecular wt. of glucose (C6H12O6) =180 gram
Wt. of Glucose (WB)= 18 gram
Wt. of water (WA)=100 gram

Q. 4. What do you mean by abnormal molecular mass ? Explain the factors with suitable examples which bring about the abnormality ?

Ans⇒ Molecular masses of the solutes which are inversely proportional to the colligative properties must have different values from their normal values. In other words, these are abnormal in nature and also called observed molecular masses.

In certain solvents (generally non-polar) the particles of the solute undergo association i.e., two or more molecules of solute combine to form a bigger molecule. As a result, the total number of

nA                                            ⇌                              (A).S
(Single molecules)                                     (Associated molecules)

Solute molecules in the solution decrease. The colligative properties are expected to decrease while the molecular masses of the solutes involved are expected to increase. It is also expected since the molecular mass associated above, varies inversely as the colligative property.

For example, in benzene solvent, both acetic acid and benzoic acid exist as dimers because of the presence of intermolecular hydrogen bonding.

2CH3COOH ⇌ (CH3COOH)2

2C6H5COOH ⇌ (C6H5COOH)2

KCl gives K+ and Cl- in aqueous solution hence the number of particles will be double in aqueous solution. There its abnormal molecular weight is half that of normal molecular weight.

Q. 5. Define vapour pressure of a liquid. What happens to the vapour pressure when (a) volatile solute dissolves in the liquid and (b) the dissolved solute is non-volatile ?

Ans⇒ Every pure liquid exerts a vapour pressure in the space above it. This is the vapour pressure of the solvent over it at that particular temperature. It depends upon the nature of the solvent and the temperature

(a) If a volatile solute is dissolved, vapour pressure of the solvent is increased.

(b) However, if a non-volatile solute is dissolved 11 it, the vapour pressure of the solution is lowered. It is because, in a solution, the percentage of the volatile solvent molecules, which only contributes towards vapour pressure is diminished.

Since, the solute molecules are non-volatile and show no measurable tendency to escape from the solute as vapour, consequently, the vapour pressure of a solution is always lower than that of its solvent.

Decrease in vapour pressure when a
a non-volatile solute is added to the solvent.

Raoult’s law gave a relation between the relative lowering of vapour pressure and the mole fraction of the solute. Mathematically :

P0 -PP0 = nn+N = mwmw+Mw    (mole fraction of the solution)

Using the above equation, we can determine the molecular weight of the solute, when the lowering in v.p. is known, when a known weight of the solute w, dissolved in a known wt. of the solvent W. P0 is the vapour pressure of the pure solvent and m and M are the molecular weights of solute and solvent respectively.

Q. 6. What is Rauolt’s law of relative lowering in vapour pressure (V.P.) ? Find out the molecular weight of non-volatile solute and also prove that molecular weight of solute is inversely proportional to lowering in V.P. ?

Ans⇒ Rauolts law of relative lowering of vapour pressure : When a non-volatile solute is added to pure solvent then vapour pressure of solution is decreased. So according to Rauolt’s law relative lowering in vapour pressure is equal to mole fraction of solute.

Let P0 is the vapour pressure of pure solvent and WB  is weight of non-volatile solute and Ps  is the vapour pressure of solution then lowering in vapour pressure is P0 -PS  or p .

Now according to Rauolt’s law

P0 -Ps P0 = XB    (Mole fraction of solute)

So , P0 -Ps P0 = XB

P0 -Ps P0 = nBnA +nB

where nA and nB are no. of moles of solvent and solute respectively.
For a very very dilute solution.

nB << nA

So , nA + nB = nA

Hence Po -PsPo= nBnA   where p = Po -Ps

ppo = nBnA

ppo = WB MAMB. WA

So , MB =WB MA Po WA . 1p

 MB = (const).1p

Hence  MB 1p

Molecular weight of solute is inversely proportional to lowering in vapor pressure (V.P.)
where WB = weight of solute taken
MB =  Molecular weight of solute
MA =molecular weight of solvent
Po -Ps = p = lowering in vapor pressure

Q. 7. State Henry’s law and mention some important applications.

Ans⇒ Henry’s Law : Effect of pressure on solubility of gasses in liquids at a particular temperature is given by Henry’s law. “The mass of a gas dissolved in a given volume of the liquid at a constant temperature is directly proportional to the pressure of the gas present in equilibrium with the liquid.”

Mathematically m ∝ p 
.                             m =kp

where m = mass of the gas dissolved in a unit volume of the solvent

p= Pressure of the gas in equilibrium with the solvent

k= Constant called Henry’s Law constant.

Some important applications of Henry’s law :

1. To increase the solubility of CO2 in soft drinks and soda-water, the bottle is sealed under high pressure.

2. Deep sea divers use oxygen diluted with less soluble helium as breathing gas to minimize the painful effects accompanying the decompression.

3. In lungs where oxygen is present in air with high partial pressure, hemoglobin combines with oxygen to form oxyhaemoglobin. In tissues where partial pressure of oxygen is low it releases oxygen for utilization in cellular activities.

Q. 8. 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⇒ The lowering of vapor pressure of freezing point compared to that of the pure solvent. The freezing point of a Substance, the solid phase is in dynamic equilibrium with the liquid phase.

The freezing point of a substance may be defined as the temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure of solid phase. A solution will freeze when its vapor pressure equals the vapor pressure of the pure solid solvent.

According to Raoult’s law, when a nonvolatile solid is added to the solvent its vapor decreases and it would become equal to that of solid solvent at lower temperature. The freezing point of the solvent decreases.

Let T0f be the freezing point of pure solvent Tf be its freezing point when non-volatile solute is dissolved in it.

Tf= T0f -Tf depression of freezing point

Depression Constant : The unit of Kf is K kg mol-1  values of Kf  for some common solvent acetic acids are listed in this way. If W2  gram of the solute having molar mass as present in W1   gram of solvent produces the depression in freezing point T1  of the solvent then molality,

The Solute is as

m= W2/M2W1 /1000 

We can find out that the freezing point for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in same order.

Q. 9. Define osmotic pressure. Prove that osmotic pressure is a molecular property.

Ans⇒ Osmotic pressure : The excess pressure exerted on the surface of solution to just prevent osmosis is called osmotic pressure. It depends upon the molal concentration. Hence it is a colligative property.

According toVan’t Hoff ∝  C (at constant temp.) and ∝  T (At constant concentration).
Both laws are combined, we have

∝  CT  and c= nv
or,   =RCT     

  =nvRT
or, u  =n RT      

or,    =WMRT     

Where M= molecular weight and R=Gas constant.

Q.10. What role does molecular interaction play in the solution of alcohol in water?
Ans⇒  In case of alcohol as well as water, the molecules are interlinked by intermolecular hydrogen bonding. However, the hydrogen bonding is also present in the molecules of alcohol and water in the solution but it is comparatively less than both alcohol and water. As a result, the magnitude of attractive forces tends to decrease and the solution shows positive deviation from Raoult’s Law. This will lead to increase in vapor pressure of the solution and also decrease in its boiling point.

Q.11. Why do gasses always tend to be less soluble in liquids as the temperature is raised?
Ans⇒  When gasses are dissolved in water, it is accompanied by a release of heat energy, i.e., process is exothermic. When the temperature is increased, according to Le Chatelier's Principle, the equilibrium shifts in backward direction, and thus gases become less soluble in liquids.

Q.11 . State Henry’s law and mention some of its important applications.
Ans⇒ Henry’s law: The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.
or
The partial pressure of a gas in vapor phase is proportional to the mole fraction of the gas (X  ) in the solution. p=KHX .
where KH  is Henry’s law constant.
Applications of Henry’s law :
(i) In order to increase the solubility of CO2  gas in soft drinks and soda water, the bottles are normally sealed under high pressure. Increase in pressure increases the solubility of a gas in a solvent according to Henry’s Law. If the bottle is opened by removing the stopper or seal, the pressure on the surface of the gas will suddenly decrease. This will cause a decrease in the solubility of the gas in the liquid i.e. water. As a result, it will rush out of the bottle producing a hissing noise or with a fiz.
(ii) As pointed above, oxygen to be used by deep sea divers is generally diluted with helium inorder to reduce or minimize the painful effects during decompression.
(iii) As the partial pressure of oxygen in air is high, in lungs it combines with hemoglobin to form oxyhaemoglobin. In tissues, the partial pressure of oxygen is comparatively low. Therefore, oxyhaemoglobin releases oxygen in order to carry out cellular activities.

Q.12. According to Raoult’s law, what is meant by positive and negative deviations and how is the sign of sol H related to positive and negative deviations from Raoult’s law?

Ans⇒  Solutions having vapor pressures more than that expected from Raoult’s law are said to exhibit positive deviation. In these solutions solvent – solute interactions are weaker and sol H is positive because stronger A-A or B-B interactions are replaced by weaker A-B interactions. Breaking of the stronger interactions requires more energy & less energy is released on formation of weaker interactions. So overall sol H is positive. Similarly sol V is positive i.e. the volume of solution is somewhat more than sum of volumes of solvent and solute.

So there is expansion in volume on solution formation.

Similarly in case of solutions exhibiting negative deviations, A-B  interactions are stronger than A-A  & B-B . So weaker interactions are replaced by stronger interactions so , there is release of energy i.e. sol H is negative.

Q.13. Suggest the most important type of intermolecular attractive interaction in the following pairs:

(i) n-hexane and n-octane

(ii) I2 and CCl4.

(iii) NaClO4 and water

(iv) methanol and acetone

(v) acetonitrile (CH3CN) and acetone (C3H6O)

Ans⇒  (i) Both w-hexane and n-octane are non-polar. Thus, the intermolecular interactions will be London dispersion forces.

(ii) Both I2 and CCl4 are non-polar. Thus, the intermolecular interactions will be London dispersion forces.

(iii) NaClO4 is an ionic compound and gives Na+ and ClO4- ions in the Solution. Water is a polar molecule. Thus, the intermolecular interactions will be ion-dipole interactions.

(iv) Both methanol and acetone are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.

(v) Both (CH3CN) and (C3H6O) are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.

Q.14. 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 (having polar -OH  group) – Partially soluble.
(ii) Toluene (non-polar) – Insoluble.
(iii) Formic acid (form hydrogen bonds with water molecules) – Highly soluble.
(iv) Ethylene glycol (form hydrogen bonds with water molecules) Highly soluble.
(v) Chloroform (non-polar)- Insoluble.
(vi) Pentanol (having polar -OH ) – Partially soluble.

Q.15 . Calculate the mass percentage of aspirin (C9H8O4 in acetonitrile (CH3CN) when 6.5 g of CHO is dissolved in 450 g of  CH3CN .

Ans⇒ 

Mass of aspirin =6.5 g

Mass of acetonitrile =450 g

Then, total mass of the solution = (6.5 +450) g =456.5 g

Therefore, mass percentage of C9H8O4 = 6.5456.5100 %   

= 1.424 %

Q.16 . 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⇒ 

Fluorine being more electronegative than chlorine has the highest electron withdrawing inductive effect. Thus, trifluoroacetic acid is the strongest trichloroacetic acid is second most and acetic acid is the weakest acid due to absence of any electron withdrawing group. Thus, F3CCOOH ionizes to the largest extent while CH3COOH ionizes to a minimum extent in water. Greater the extent of ionization is the depression in freezing point. Hence, the order of depression in freezing point will be CH3COOH < Cl3CCOOH <F3CCOOH .

Q.17. Define the terra solution. How many types of solutions are formed? Write briefly about each type with an example.

Ans⇒  A solution is a homogeneous mixture of two or more chemically non-reacting substances. Types of solutions: There are nine types of solutions.

Types of Solution Examples

Gaseous solutions

(a) Gas in gas Air, mixture of O2 and N2 , etc.

(b) Liquid in gas Water vapor

(c) Solid in gas Camphor vapors in N2 gas, smoke etc.

Liquid solutions

(a) Gas in liquid CO2 dissolved in water (aerated water), and O2 dissolved in water, etc.

(b) Liquid in liquid Ethanol dissolved in water, etc.

(c) Solid in liquid Sugar dissolved in water, saline water, etc.

Solid solutions

(a) Gas in solid Solution of hydrogen in palladium

(b) Liquid in solid Amalgams, e.g., Na-Hg

(c) Solid in solid Gold ornaments (Cu/Ag with Au)

Q.18 . Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.

Ans⇒  0.25 Molal aqueous solution to urea means that

moles of urea =0.25 mole

mass of solvent (NH2CONH2) = 60 g mol-1

.’. 0.25 mole of urea =0.25 60 = 15 g

Mass of solution = 1000 +15 =1015 g = 1.015 kg

1.015 kg of urea solution contains 15 g of urea

.’. 2.5 kg of solution contains urea = 151.0152.5 =37 g

Q.19 . Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Ans⇒  Mass of solution = Mass of C6H6 + Mass of CCl4

= 22 g + 122 g = 144 g

Mass % of benzene = 22144 100 = 15.28 %  

Mass % of CCl4 = 122144 100 = 84.72% 

Q.20 .  A solution of glucose in water is labeled as 10 % w/w , that would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g/ml then what shall be the molarity of the solution?

Ans: We have given that,

10 % w/w solution of glucose in water i.e., 10 g of glucose in 90 g of water.

Now, as we know;

Molar mass of glucose =180 g/mol

Molar mass of water =18 g/mol

Thus, 

Number of moles of glucose in the solution =10180 = 0.055 mol

Number of moles of water in the solution =9018 = 5 mol

Taking into consideration the above values;

Molality is given as – 

Molality = 0.055 100090 =0.617 m

Mole fraction of each component can be given as – 

Mole fraction of glucose = 0.0550.055 + 5 =0.0108

Mole fraction of water =1 - 0.0108=0.9892

Again,

We have given, density of solution is 1.2 g/ml . Thus,

Volume =massdensity = 1001.2 = 83.33 ml

Thus,

Molarity can be given as;

Molarity = 0.055 100083.33 = 0.66 M

Q.21. A solution is obtained by mixing 300 g of 25 % solution and 400 g of 40 % solution by mass. Calculate the mass percentage of the resulting solution.

Ans⇒ Mass percentage is defined as the ratio of mass of solute by the mass of solution multiplied by 100 .

Mass percentage = massolutemasssolution100

Here,

We have given a mixture of 300 g of 25 % solution and 400 g of 40 % solution by mass. Thus,

300 g of 25 % solution will contain = 25 300100 = 75 g of solute

400 g of 40 % solution will contain = 40 400100 = 160 g of solute

Now, the resulting solution will have contents as;

Total mass of solute =75 +160 = 235 g  

Total mass of solution =300 +400 = 700 g

Thus,

Mass percentage is given as,

Mass percentage of solute in the solution =235700 100 =33.57 %

Similarly,

Mass percentage of water in the solution =100-33.57 = 66.42 %

Q.22. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water.

(1) Phenol

Ans⇒ Insoluble in water:

(2) Toluene

Ans⇒ Toluene is not soluble in one another due to their non-polar nature.

(3) Formic acid

Ans⇒ Partially soluble in water:

(4) Ethylene glycol

Ans⇒ Partially soluble

(5) Chloroform 

Ans⇒Highly insoluble in water:

(6) Pentanol

Ans⇒Partially soluble in water due to the presence of the alcohol functional group.

Q.23. If the solubility product of CuS is 6 10-16 , calculate the maximum molarity of CuS aqueous solution.

Ans⇒ The dissociation reaction is given as;

CuS ⇌ Cu2+ + S2-

The solubility product is given as; Ksp = 6 10-16

Let the solubility of each ion be denoted as X mol/L .

Thus,

Ksp = [Cu2+] [S2-] = X X

X2 = 6 10-16

= 6 10-16 

∴ Maximum molarity of CuS,

X = 2.44 10-8 mol/L

Q.24 . Determine the amount of CaCl2 ( i=2.47 ) dissolved in 2.5 liters of water such that its osmotic pressure is 0.75 atm at 27 C .

Ans⇒ Given that,

Volume = 2.5 L

Van’t Hoff factor =2.47

Osmotic pressure =0.75 atm

Gas constant = 0.0821 Latm K-1 mol-1

Temperature =273 +27 =300 K

Molar mass of CaCl2 = 111 g/mol

Osmotic pressure is given as,

= i CRT

= i  nV RT = i WM V RT

W = M V i RT = 0.75 111 2.52.47 0.0821300

Thus, the amount of CaCl2 required,

W = 3.42 g

Q.25 . At 300 K , 36 g of glucose present in a liter of its solution has an osmotic pressure of 4.08 bar . If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Ans⇒ The formulation for the osmotic pressure is given as;

= CRT

Case 1 – 

T = 300 K

Mass of glucose =36 g

Molar mass of glucose =180 g/mol

Osmotic pressure =4.08 bar

According to the formula;

= WM RT

4.08 = 36180 R 300

 R= 0.068 units

Case 2 – 

Osmotic pressure =1.52 bar

T = 300 K

Thus, by given formula;

= CRT

Thus, the concentration will be,

C = 1.520.068 300 = 0.0745 M

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