NCERT Grade 12-Solutions-Answers

NCERT Solutions for Class 12 Chemistry

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1.       

Alternatively, Mass percentage of CCl4 = (100 – 15.28) %

= 84.72%

2.    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 (C6H6) = (6 ×  12 + 6 ×) g mol-1

= 78 g mol-1

∴ Number of moles of 

= 0.3846 mol

Molar mass of carbon tetrachloride (CCl) = 1 × 12 + 4 × 355

= 154 g mol-1

∴ Number of moles of 

= 0.4545 mol

Thus, the mole fraction of C6His given as:

3.    Molarity is given by:

(a) Molar mass of Co(NO3)2 = 59 + 2 (14 + 3 × 16) + 6 × 18

= 291 g mol-1

Therefore, Moles of Co(NO3)2 = 30/291 mol

= 0.103 mol

Therefore, molarity 

= 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol

∴ Number of moles present in 30 mL of 0.5 M 

= 0.015 mol

Therefore, molarity 

= 0.03 M

4.     Molar mass of urea NH2CONH2 = 2 (1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g mol-1

0.25 molar aqueous solution of urea means:

1000 g of water contains 0.25 mol = (0.25 × 60) of urea

= 15 g of urea

That is,

(1000 + 15) g of solution contains 15 g of urea

Therefore, 2.5 kg (2500 g) of solution contains 

= 36.95 g= 37 g of urea (approximately)

Hence, mass of urea required = 37 g.

5.    (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 = 1.202 g ml-1

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.    It is given that the solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of  H2S is dissolved in 1000 g of water.

Moles of water = 

= 55.56 mol

∴Mole fraction of H2S, x = 

= 0.0035

At STP, pressure (p) = 0.987 bar

According to Henry’s law:

p= KHx

= 282 bar.

7.    It is given that:

KH = 1.67 × 108 Pa

PCO2 = 2.5 atm = 2.5 × 1.01325 × 105 Pa

= 2.533125 × 105 Pa

According to Henry’s law:

= 0.00152

We can write, 

[Since, nCO2 is negligible as compared to nH2O]

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 COin 500 mL of soda water = (0.042 × 44) g

= 1.848 g.

8.   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, ptotal = pA + pB

Therefore, xB = 1 – xA

= 1 – 0.4

= 0.6

Now, 

= 450 × 0.4

= 180 mm of Hg

= 700 × 0.6

= 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.    It is given that vapour pressure of water,   of Hg

Weight of water taken, w1 = 850 g

Weight of urea taken, w2 = 50 g

Molecular weight of water, M1 = 18 g mol-1

Molecular weight of urea, M2 = 60 g mol-1

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.  Here, elevation of boiling point ΔTb = (100 + 273) – (99.63 + 273)

= 0.37 K

Mass of water, w1 = 500 g

Molar mass of sucrose (C12H12O11), = M2 = 11 × 12 + 22 × 1 + 11 × 16

= 342 g mol-1

Molal elevation constant, Kb = 0.52 K kg mol-1

We know that:

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.

11.   Mass of acetic acid, w1 = 75 g

Molar mass of ascorbic acid (C6H8O6), = M2 = 6 × 12 + 8 × 1 + 6 × 16

= 176 g mol-1

Lowering of melting point, ΔTf = 1.5 K

We know that:

= 5.08 g (approx)

Hence, 5.08 g of ascorbic acid is needed to be dissolved.

12.  It is given that:

Volume of water, V= 450 mL = 0.45 L

Temperature, T = (37 + 273) K = 310 K

Number of moles of the polymer, 

We know that:

Osmotic pressure, 

= 30.98 Pa

= 31 Pa (approximately).

13.  Homogeneous mixtures of two or more than two components are known as solutions.

There are three types of solutions.

(i) Gaseous solution: The solution in which the solvent is a gas is called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.

(ii) Liquid solution: The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas, liquid, or solid. For example, a solution of ethanol in water is a liquid solution.

(iii) Solid solution: The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution

14.  In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid solution will be formed. For example, a solution of hydrogen in palladium is a solid solution in which the solute is a gas.

15.  (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 = 

16.  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 (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol-1

Then, number of moles of 

= 1.079 mol

Given,

Density of solution = 1.504 g mL-1

Therefore, Volume of 100 g solution = 

= 66.49 mL

= 66.49 × 10-3 L

Molarity of solution = 

= 16.23 M

17.  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 (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol-1

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 xw = 1 – xg

= 1 – 0.011

= 0.989

If the density of the solution is 1.2 g ml-1, then the volume of the 100 g solution can be given as:

= 83.33 mL

= 83.33 × 10-3 L

∴ Molarity of the solution = 

= 0.67 M

18.  Let the amount of Na2CO3 in the mixture be x g.

Then, the amount of NaHCO3 in the mixture is (1 – x) g.

Molar mass of Na2CO3 = 2 × 23 + 1 × 12 + 3 × 16

= 106 g mol-1

∴ Number of moles Na2CO3

Molar mass of NaHCO= 1 × 23 + 1 × 1 × 12 + 3 × 16

= 84 g mol-1

Number of moles of NaHCO

According to the question,

⇒ 84x = 106 – 106x

⇒ 190x = 106

⇒ x = 0.5579

Therefore, number of moles of Na2CO3

= 0.0053 mol

And, number of moles of NaHCO3 =

= 0.0053 mol

HCl reacts with Na2CO3 and NaHCO3 according to the following equation.

1 mol of Na2CO3 reacts with 2 mol of HCl.

Therefore, 0.0053 mol of Na2CO3 reacts with 2 × 0.0053 mol = 0.0106 mol.

Similarly, 1 mol of NaHCO3 reacts with 1 mol of HCl.

Therefore, 0.0053 mol of NaHCO3 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 Na2CO3 and NaHCO3, containing equimolar amounts of both.

19.  Total amount of solute present in the mixture is given by,

= 75 + 160

= 235 g

Total amount of solution = 300 + 400 = 700 g

Therefore, mass percentage (w/w) of the solute in the resulting solution, 

= 33.57%

And, mass percentage (w/w) of the solvent in the resulting solution,

= (100 – 33.57) %

= 66.43%.

20.  Molar mass of ethylene glycol [C2H4(OH)2] = 2 × 12 + 6 × 1 + 2 × 16

= 62 g mol-1

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

= 0.3942 × 10-3 L

Molarity of the solution = 

= 9.11 M

21.  (i) 15 ppm (by mass) means 15 parts per million (106) of the solution.

Therefore, percent by mass = 

= 1.5 × 10-3 %

(ii) Molar mass of chloroform (CHCl3) = 1 × 12 + 1 × 1 + 3 × 35.5

= 119.5 g mol-1

Now, according to the question,

15 g of chloroform is present in 106 g of the solution.

i.e., 15 g of chloroform is present in (106 – 15) Ãc‰ˆ 106 g of water.

Therefore, Molality of the solution = 

= 1.26 × 10-3 m.

22.  In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol-alcohol and water-water interactions. As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.

23.  Solubility of gases in liquids decreases with an increase in temperature. This is because dissolution of gases in liquids is an exothermic process.

Gas + Liquid → Solution + Heat

Therefore, when the temperature is increased, heat is supplied and the equilibrium shifts backwards, thereby decreasing the solubility of gases.

24.  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:

p = KHx

Where, Kis 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 to 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.

25.  Molar mass of ethane (C2H6) = 2 × 12 + 6 × 1

= 30 g mol-1

Number of moles present in 6.56 × 10-3 of ethane = 

= 2.187 × 10- 4 mol

Let the number of moles of the solvent be x.

According to Henry’s law,

p = KHx

1 bar 

1 bar

25.  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, ΔSolH = Positive

In the case of solutions showing negative deviations, evolution of heat takes place.

Therefore, ΔSolH = Negative.

27.  Here,

Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar

Vapour pressure of pure water at normal boiling point (p10) = 1.013 bar

Mass of solute, (w2) = 2 g

Mass of solvent (water), (w1) = 98 g

Molar mass of solvent (water), (M1) = 18 g mol-1

According to Raoult’s law,

Hence, the molar mass of the soulte is 41.35 g mol-1

28.  Vapour pressure of heptanes 

Vapour pressure of octane  = 46.8 kPa

We know that,

Molar mass of heptane (C7H16) = 7 × 12 + 16 × 1

= 100 g mol-1

Therefore, Number of moles of heptane = 

= 0.26 mol

Molar mass of octane = (C8H18) = 8 × 12 + 18 × 1

= 114 g mol-1

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,

= 0.456 × 105.2

= 47.97 kPa

Partial pressure of octane,

= 0.544 × 46.8

= 25.46 kPa

Hence, vapour pressure of solution, ptotal = p1 + p2

= 47.97 + 25.46

=73.43 kPa.

29.  1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).

Molar mass of water = 18 g mol-1

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 – p1 = 0.2177

⇒ p1 = 12.0823

Hence, the vapour pressure of the solution is 12.08 kPa.

30.  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, M2 = 40 g mol-1

Mass of octane, w1 = 114 g

Molar mass of octane, (C8H18), M1 = 8 × 12 + 18 × 1

= 114 g mol-1

Applying the relation,

Hence, the required mass of the solute is 8 g.

31.  (i) Let, the molar mass of the solute be M g mol-1

Now, the no. of moles of solvent (water), 

And, the no. of moles of solute, 

p1 = 2.8 kpa

Applying the relation:

After the additin 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 23 g mol-1

(ii) Putting the value of Ã¢â‚¬ËœM’ in equation (i), we have:

Hence, the vapour pressure of water at 298 K is 3.53 kPa.

32.  Here, ΔT= (273.15 – 271) K

= 2.15 K

Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16

= 342 g mol-1

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, 

= 0.1537 mol kg-1

Applying the relation,

Molar of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16

= 180 g mol-1

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, 

= 0.2926 mol kg-1

Applying the relation,

ΔT= Kf × m

= 13.99 K  kg mol-1 × 0.2926 mol kg-1

= 4.09 K (approximately)

Hence, the freezing point of 5% glucose solution is (273.15 – 4.09) K= 269.06 K.

33.  We know that,

Now, we have the molar masses of AB2 and AB4 as 110.87 g mol-1 and 196.15 g mol-1 respectively.

Let the atomic masses of A and B be x and y respectively.

Now, we can write:

x + 2y = 110.87     ..(i)

x + 4y = 4y = 196.15   ..(ii)

Subtracting equation (i) from (ii), we have

2y = 85.28

⇒ y = 42.64

Putting the value of ‘y‘ in equation (1), we have

x + 2 × 42.64 = 110.87

⇒ x = 25.59

Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.

34.  Here,

T = 300 K

π = 1.52 bar

R = 0.083 bar L K-1 mol-1

Applying the relation,

π = CRT

= 0.061 mol

Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.

35.  (i) Van der Wall’s forces of attraction.

(ii) Van der Wall’s forces of attraction.

(iii) Ion-diople interaction.

(iv) Dipole-dipole interaction.

(v) Dipole-dipole interaction.

36.  n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-octane.

The order of increasing polarity is:

Cyclohexane < CH3CN < CH3OH < KCl

Therefore, the order of increasing solubility is:

KCl < CH3OH < CH3CN < Cyclohexane.

37.  (i) Phenol (C6H5OH) has the polar group -OH and non-polar group-C6H5. Thus, phenol is partially soluble in water.

(ii) Toluene (C6H5 – CH3) 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 (C5H11OH) has polar -OH group, but it also contains a very bulky non-polar ­­­-C5H11 group. Thus, pentanol is partially soluble in water.

38.  Number of moles present in 92 g of Na+ ions =  = 4 mol

Therefore, molality of Na+ ions in the lake   = 4 m.

39.  Solubility product of CuS, Ksp = 6 × 10-16

Let s be the solubility of CuS in mol L.

Now, Ksp = [Cu2+] [S2-]

= s × s

= s2

Then, we have, Ksp = s2 = 6 × 10-16

= 2.45 × 10 – 8 mol L-1

Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10-8 mol L-1

40.  6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

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

= 456.5 g

Therefore, mass percentage of 

= 1.424%.

41.  The molar mass of nalorphene C19H21NO3 is given as:

19 × 12 + 21 × 1 + 1 × 14 + 3 × 16 = 311 g mol-1

In 1.5 × 10-3 m aqueous solution of nalorphene,

1 kg (1000 g) of water contains 1.5 × 10-3 mol = 1.5 × 10-3 × 311 g

= 0.4665 g

Therefore, total mass of the solution = (1000 + 0.4665) g

= 1000.4665 g

This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.

Therefore, mass of the solution containing 1.5 mg of nalorphene is:

= 3.22 g

Hence, the mass of aqueous solution required is 3.22 g.

42.  0.15 M solution of benzoic acid in methanol means,

1000 mL of solution contains 0.15 mol of benzoic acid

Therefore, 250 mL of solution contains  mol of benzoic acid

= 0.0375 mol of benzoic acid

Molar mass of benzoic acid (C6H5COOH) = 7 × 12 + 6 × 1 + 2 × 16

= 122 g mol-1

Hence, required benzoic acid = 0.0375 mol × 122 g mol-1

= 4.575 g.

43.  

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 H+ 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

44.  Molar mass of CH3CH2CHClCOOH = 15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1

= 122.5 g mol-1

Therefore, No. of moles present in 10 g of 

= 0.0816 mol

It is given that 10 g of CH3CH2CHClCOOH is added to 250 g of water.

Therefore, Molality of the solution, 

= 0.3264 mol kg-1

Let a be the degree of dissociation of CH3CH2CHClCOOH

CH3CH2CHClCOOH undergoes dissociation according to the following equation:

Initial conc. At equilibrium 

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

= 0.0655

Again,

Initial moles at equilibrium  

Total moles of equilibrium = 1 – α + α + α

= 1 + α

Therefore, 

= 1 + 0.0655

= 1.0655

Hence, the depression in the freezing point of water is given as:

ΔTf = i.Kfm

= 1.0655 × 1.86 K kg mol-1 × 0.3264 mol-1 kg

= 0.65 K.

45.  It is given that:

Therefore, observed molar mass of CH2FCOOH, (M2)obs = 72.54 g mol

The calculated molar mass of CH2FCOOH is:

(M2)cal = 14 + 19 + 12 + 16 + 16 + 1 = 76 g mol

Therefore, van’t Hoff factor, 

= 1.0753

Let αbe the degree of dissociation of CH2FCOOH

Initial conc. At equilibrium 

= 1.0753 – 1

= 0.0753

Now, the value of Ka is given as:

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

46.  Vapour pressure of water,  = 17.535 mm of Hg

Mass of glucose, w2 = 25 g

Mass of water, w1 = 450 g

We know that,

Molar mass of glucose (C6H12O6), M2 = 6 × 12 + 12 × 1 + 6 × 16

= 180 g mol-1

Molar mass of water, M1 = 18 g mol-1

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.

47.  Here,

p = 760 mm Hg

kH = 4.27 × 105 mm Hg

According to Henry’s law,

Hence, the mole fraction of methane in benzene is 178 × 10-5.

48.  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, xB = 1 – 0.114

= 0.886

Vapour pressure of pure liquid B,  = 500 torr

Therefore, vapour pressure of liquid B in the solution,

= 500 × 0.886

= 443 torr

Total vapour pressure of the solution, ptotal = 475 torr

Therefore, Vapour pressure of liquid A in the solution,

Hence, the vapour pressure of pure liquid A is 280.7 torr.

49.  From the question, we have the following data

It can be observed from the graph that the plot for the total of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

50.  Molar mass of benzene (CH) = 6 × 12 + 6 × 1

= 78 g mol

Molar mass of toluene (CHCH) = 7 × 12 + 8 × 1

= 92 g mol

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, xt = 1 – 0.486

= 0.514

It is given that vapour pressure of pure benzene, pb0 = 50.71 mm Hg

And, vapour pressure of pure toluene, p10 = 32.06 mm Hg

Therefore, partial vapour pressure of benzene, pb = xb × pb

= 0.486 × 50.71

= 24.645 mm Hg

And, partial vapour pressure of toluene, p1 = xt × pt

= 0.514 × 32.06

= 16.479 mm Hg

Hence, mole fraction of benzene in vapour phase is given by:

= 0.599

= 0.6

51.  Percentage of oxygen (O2) in air = 20 %

Percentage of nitrogen (N2) in air = 79%

Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, (10 × 760) 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:

p = KHx For oxygen:

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10-5 and 9.22 × 10-5 respectively.

52.  We know that,

V = 2.5 L

i = 2.47

T = (27 + 273)K = 300 K

Here,

R = 0.0821 L atm K-1 mol-1

M = 1 × 40 ÷ 2 × 35.5

= 111 g mol-1

Therefore, 

= 3.42 g

Hence, the required amount of CaCl2 is 3.42 g.

53.  When K2SO4 is dissolved in water, K+ and SO42- ions are produced.

Total number of ions produced = 3

Therefore, i = 3

Given,

w = 25 mg = 0.025 g

V = 2 L

T = 250 C = (25 + 273) K = 298 K

Also, we know that:

R = 0.0821 L atm

M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol-1

Applying the following relation,

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