//M3//QN1//SUB//DL0

Define the term solution. How many types of solutions exist? Write briefly about each type with an example.

//X

//M1//QN2//SUB//DL0

Give an example of a solid solution in which the solute is a gas.

//X

//M4//QN3//SUB//DL0

Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass Percentage

//X

//M2//QN4//SUB//DL0//EQ

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is

1.504 g.mL–1?

//X

d = 1.504 g.mL–1, molarity (M) = ?

68% nitric acid means that 68 g of nitric acid is dissolved in 100 g of solution.

Molar mass of HNO3 = 63 g.mol–1

Volume of Solution =

=

= 66.489 mL

= 0.0665 L

Molarity

=

=

= 16.23 M

//M4//QN5//SUB//DL0//EQ

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is

1.2 g.mL–1, then what shall be the molarity of the solution?

//X

10% W/W glucose solution means that

mass of glucose = 10 g, mass of H2O = 90 g

Molar mass of glucose = 180 g.mol–1

(1) Molality (m):

m =

=

= 0.617 m

(2) Mole-fraction:

Moles of glucose = = 0.055 mol

Moles of water = = 5 mol

Total Moles = 5 + 0.055

= 5.055

Mole-fraction of glucose = = 0.0108

Mole-fraction of water = 1 – 0.0108

= 0.989

(3) Molarity (M):

Volume of Solution =

=

= 83.33 mL

= 0.0833 L

M =

=

= 0.67 M

//M4//QN6//SUB//DL0//EQ

How many ml of 0.1 M HC1 are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?

//X

Let, mass of Na2CO3 = x g

∴ Mass of NaHCO3 = (1 – x) g

Molar mass of Na2CO3 = 106 g.mol–1

Molar mass of NaHCO3 = 84 g.mol–1

Na2CO3 and NaHCO3 both are equimolar

∴ Moles of Na2CO3 = Moles of NaHCO3

∴ =

∴ 84x = 106 – 106x

∴ 84x + 106x = 106

∴ 190x = 106

∴ x = 0.5578 g

∴ Mass of Na2CO3 = x = 0.5578 g

∴ Mass of NaHCO3 = 1 – x

= 1 – 0.5578

= 0.4422 g

Let us find out mass of HCl which react with Na2CO3 and NaHCO3

Na2CO3 + 2 HCl → 2 NaCl + H2O + CO2

106 g = 2 × 36.5

= 73 g

If 106 g Na2CO3 react with 73 g of HCl

Then 0.5578 g Na2CO3 react with:

Mass of HCl =

= 0.384 g

NaHCO3 + HCl → NaCl + CO2 + H2O

84 g 36.5 g

If 84 g NaHCO3 react with 36.5 g of HCl

Then 0.4422 g NaHCO3 react with:

Mass of HCl =

= 0.192 g

Total mass of HCl = 0.384 + 0.192

= 0.576 g

M =

∴ 0.1 =

∴ Volume of solution =

= 0.1578 L

= 157.8 mL

//M2//QN7//SUB//DL0//EQ

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.

//X

Ans. Solution 1:

Mass of Solute =

=

= 75 g

Solution 2:

Mass of Solute =

=

= 160 g

Total mass of solute = 75 + 160

= 235 g

Total mass of solution = 300 + 400

= 700 g

∴ Mass Percentage of Resulting Solution

=

= 33.57%

//M3//QN8//SUB//DL0//EQ

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?

//X

Molar mass of ethylene glycol = 62 g

molality =

= 17.95 m

Mass of ethylene glycol = 222.6 g

Mass of water = 200.0 g

Total mass of solution = 422.6 g

Density =

\ Volume of solution =

= 394.22 mL

= 0.394 L

Molarity (M)

=

=

= 9.11 M

//M3//QN9//SUB//DL0//EQ

A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass (ii) determine the molality of chloroform in the water sample.

//X

15 ppm (by mass) CHCl3 means that 15 g CHCl3 present in 106 g of solution.

(i) %W/W = × 100

=

= 1.5 × 10–3 %

(ii) Molality:

m =

=

= 1.25 × 10–4 m

//M2//QN10//SUB//DL0

What role does the molecular interaction play in a solution of alcohol and water?

//X

//M0//QN11//SUB//DL0

Why do gases always tend to be less soluble in liquids as the temperature is raised?

//X

//M0//QN12//SUB//DL0//EQ

State Henry's law and mention some important applications.

//X

//M0//QN13//SUB//DL0//EQ

The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar.

If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas ?

//X

Molar mass of C2H6 (ethane) = 30 g.mol–1

Moles of ethane =

= 0.218 × 10–3

= 2.18 × 10–4

Let, moles of solvent = x

According to Henry's Law

P = KH .

\ P = KH .

Moles of Solute (C2H6) << Moles of Solvent

\ 1 = KH .

\ KH =

Moles of ethane =

= 0.166 × 10–2

= 1.66 × 10–3

P = KH . x

= .

= 0.761 × 10–3 + 4

P = 7.61 bar

Second Method:

6.56 × 10–3 g of ethane → 1 bar

\ 5.00 × 10–2 g of ethane → (?)

P = = 7.62 bar

//M4//QN14//SUB//DL0

What is meant by positive and negative deviations from Raoult's law and how is the sign of DmixH related to positive and negative deviations from Raoult's law?

//X

//M2//QN15//SUB//DL0//EQ

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

//X

2% non-volatile solution means

w2 = 2 g M2 = ?

w1 = 98 g M1 = 18 g.Mol–1

p1 = 1.004 bar = 1.013 bar

=

=

\ =

\ M2 =

= 41.35 g/mol–1

//M3//QN16//SUB//DL0//EQ

Heptane and octane form an ideal solution. At 373K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g

of octane?

//X

heptane = 1 octane = 2

= 105.2 kPa = 46.8 kPa

w1 = 26.0 g w2 = 35 g

pTotal = ?

Molar mass of heptane (M1) = 7(C) + 16(H)
= 7(12) + 16(1)
= 84 + 16 = 100 g.mol–1

Molar mass of octane (M2) = 8(C) + 18(H)

= 8(12) + 18(1) = 96 + 18

= 114 g.mol–1

Moles of heptane (n1) =

=

= 0.26 Mol

Moles of octane (n2) =

=

= 0.31 Mol

Mole-fraction of heptane (x1) =

= 0.456

Mole-fraction of octane (x2) = 1 – 0.456

= 0.544

According to Raoult's Law

pTotal = p1 + p2

= . x1 + . x2

= (105.2)(0.456) + (46.8)(0.544)

= 47.97 + 25.459

= 73.43 K Pa

//M2//QN17//SUB//DL0//EQ

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

//X

= 12.3 kPa

p1 = ?

1 molal solution means that 1 mole of solute in 1000 g of solvent

n2 = 1, n1 = = 55.55

=

=

\ 12.3 – p1 =

\ 12.3 – p1 = 0.2175

\ p1 = 12.3 – 0.2175

\ p1 = 12.08 K Pa

//M2//QN18//SUB//DL0//EQ

Calculate the mass of a non-volatile solute (molar mass 40 g.mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

//X

Vapour pressure of pure octane be

Vapour pressure of solution =

= 0.8

w2 = ? M2 = 40 g.mol–1

w1 = 114 g M1 = (8 × 12) + (18 × 1)

= 114 g.mol–1

=

=

//M4//QN19//SUB//DL0//EQ

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298K. Further,

18 g of water is then added to the solution and the new vapour pressure becomes

2.9 kPa at 298K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298K.

//X

w2 = 30 g w1 = 90 g

M2 = ? p1 = 2.8 kPa

= ?

=

\ =

\ 1 – =

\ 1 – =

\ 1 – =

\ 1 – =

\ 1 – =

\ =

\ = ...(1)

After adding 18 g of water

=

\ 1 – =

\ 1 – =

\ 1 – =

\ 1 – =

\ 1 – =

\ =

\ = .....(2)

Now ratio of eq. (1) & (2)

=

\ =

\ (5 + M) 2.9 = (6 + M) 2.8

\ 14.5 + 2.9 M = 16.8 + 2.8 M

\ 2.9 M – 2.8 M = 2.3

\ 0.1 M = 2.3

\ M = 23 g/Mol

Substituting value of M in eq. (1)

=

\ =

\ =

= 3.53 kPa

//M3//QN20//SUB//DL0//EQ

A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15K.

//X

Cane Sugar: 5% W/W

w2 = 5 g w1 = 95 g

M2 = 342 g/mol Tf = 271K, = 273.15K

DTf = – Tf

= 273.15 – 271

= 2.15K

DTf = Kf .

\ Kf =

=

Kf = 13.97 K.kg.mol–1

Glucose: 5% W/W

w2 = 5g w1 = 95 g

M2 = 180 g.mol–1 DTf = ?

DTf = Kf .

= 13.97 ×

DTf = 4.08K

\ DTf = – Tf

\ 4.08 = 273.15 – Tf

\ Tf = 273.15 – 4.08

= 269.06K

//M4//QN21//SUB//DL0//EQ

Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3K whereas 1.0 g of AB4 lowers it by 1.3K. The molar depression constant for benzene is 5.1 K.kg.mol–1. Calculate atomic masses of

A and B.

//X

M2 = Kf .

For AB2: w2 = 1 g w1 = 20 g

= ? DTf = 2.3K

Kf = 5.1 K.kg.mol–1

MAB2 = = 110.87 g.mol–1

For AB4: w2 = 1 g w1 = 20 g

DTf = 1.3K MAB4 = ?

MAB4 =

= 196.15 g.mol–1

A + 2B = 110.87

A + 4B = 196.15

– – –

– 2B = – 85.28

\ B = 42.64 u

Putting value of B in eq.

A + 2B = 110.87

A + 2(42.64) = 110.87

A + 85.28 = 110.87

A = 25.59 u

//M2//QN22//SUB//DL0//EQ

At 300K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

//X

T = 300K, w2 = 36 g, M2 = 180 g.mol–1

p = 1.52 bar, C = ?

p = CRT

1.52 = C × 0.083 × 300

\ C =

= 0.0602 Mol/L

//M1//QN23//SUB//DL0

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).

//X

//M0//QN24//SUB//DL0

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.

//X

//M0//QN25//SUB//DL0

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

//X

//M2//QN26//SUB//DL0//EQ

If the density of some lake water is 1.25

g.mL–1 and contains 92 g of Na+ ions per kg

of water, calculate the molarity of Na+ ions in the lake.

//X

Molality =

=

= 4 m or mol.kg–1

//M2//QN27//SUB//DL0//EQ

If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.

//X

CuS(s) +

S S

Solubility

KSP = [Cu+2] . [S–2]

\ 6 × 10–16 = S . S

\ 6 × 10–16 = S2

\ S =

= 2.45 × 10–8 mol.L–1

//M2//QN28//SUB//DL0//EQ

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

//X

Mass of aspirin = 6.5 g

Mass of acetonitrile = 450 g

Mass of solution = 456.5 g

% W/W = = 1.42%

//M2//QN29//SUB//DL0//EQ

Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10–3 m aqueous solution required for the above dose.

//X

Molality = 1.5 × 10–3 m,

Mass of Nalorphene = 1.5 mg

= 1.5 × 10–3 g

Mass of solvent = ?

Molar mass of Nalorphene = 311 g.mol–1

m =

\ 1.5 × 10–3 =

\ Mass of solvent = 0.00321 kg = 3.21 g

//M2//QN30//SUB//DL0

Calculate the amount of benzoic acid (C6H5COOH) required for preparing

250 mL of 0.15 M solution in methanol.

//X

//M2//QN31//SUB//DL0

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.

//X

//M4//QN32//SUB//DL0//EQ

Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH

is added to 250 g of water, Ka = 1.4 × 10–3,

Kf = 1.86 K.kg.mol–1.

//X

w2 = 10 g, w1 = 250 g

M2 = 4(C) + 7(H) + Cl + 2(O) Ka = 1.4 × 10–3

= 4(12) + 7(1) + 35.5 + 2(16)

= 122.5 g.mol–1 Kf = 1.86 K.kg.mol–1

DTf = ?

Molality

C =

=

= 0.3265 Mol/L

Ka =

1.4 × 10–3 = ∝2 . (0.3265)

4.287 × 10–3 = ∝2

0.4287 × 10–2 = ∝2

∝ = 0.6547 × 10–1

∝ = 0.06547

∝ =

0.06547 =

i = 1.06547

So, depression in the freezing point is

DTf = i . Kf . m

= (1.06547)(1.86)(0.3265)

= 0.647K

//M4//QN33//SUB//DL0//EQ

19.5 g of CH2FCOOH is dissolved in

500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the Van't Hoff factor and dissociation constant of fluoroacetic acid.

//X

w2 = 19.5 g w1 = 500 g

M2 = 78 g.mol–1 DTf = 1.0°C

Kf = 1.86 K.kg.mol–1

i = ? Ka = ?

DTf = i . Kf .

\ i =

=

\ i = 1.07526

∝ =

\ ∝ =

\ ∝ = 0.07526

Molality =

=

= 0.5131

Ka =

=

=

= 0.00313

Ka = 3.1 × 10–3

//M2//QN34//SUB//DL0//EQ

Vapour pressure of water at 293K is

17.535 mm Hg. Calculate the vapour pressure of water at 293K when 25 g of glucose is dissolved in 450 g of water.

//X

w2 = 25 g w1 = 450 g

M2 = 180 g.mol–1 M1 = 18 g.mol–1

= 17.535 p1 = ?

n1 = = = 25

n2 = = = 0.14

\ =

\ =

\ 17.535 – p1 =

\ 17.535 – p1 = 0.0976

\ p1 = 17.535 – 0.0976

= 17.44 mm Hg

//M2//QN35//SUB//DL0//EQ

Henry’s law constant for the molality

of methane in benzene at 298K is

4.27 × 105 mm Hg. Calculate the solubility

of methane in benzene at 298K under

760 mm Hg.

//X

KH = 4.27 × 105 mm Hg

P = 760 mm Hg

Solubility (X) = ?

According to Henry's Law

P = KH . X

\ X =

=

= 177.98 × 10–5

\ X = 1.78 × 10–3

//M3//QN36//SUB//DL0//EQ

100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

//X

Liquid : A Liquid : B

WA = 100 g WB = 1000 g

MA = 140 g.Mol–1 MB = 180 g.Mol–1

p0A = (?) p0B = 500 torr

pA = (?)

pTotal = 475 torr

Moles of liquid–A (nA) = = = 0.714

Moles of liquid–B (nB) = = = 5.55

Mole-fraction of liquid : A (xA) =

= 0.114

Mole-fraction of liquid : B (xB) = 1 – 0.114

= 0.886

pTotal = p0A . xA + p0B . xB

475 = p0A . (0.114) + (500)(0.886)

\ 475 = p0A . (0.114) + 443

\ 475 – 443 = p0A . (0.114)

\ 32 = p0A . (0.114)

\ p0A = 280.7 torr

Vapour pressure of pure liquid A is 280.7 torr

Vapour pressure of liquid A gaseous in solution

pA = p0A . xA

= (280.7) . (0.114)

pA = 32 torr

//M4//QN37//SUB//DL0//EQ

Vapour pressures of pure acetone and chloroform at 328K are 741.8 mm Hg and 632.8 mmHg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is: 100 X xacetone 0 11.8 23.4 36.0 50.8 58.2 64.5 72.1 pacetone/mm Hg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1 pchloroform/mm Hg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7 Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

//X

In the graph ptotal of the solution curves downwards. Therefore the solution shows negative deviation from the ideal behaviour.

//M4//QN38//SUB//DL0//EQ

Benzene and toluene from ideal solution over

the entire range of composition. The vapour pressure of pure benzene and toluene at 300K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

32.1 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 78 gm of benzene is mixed with 138 gm of toluene.

//X

Ans. Benzene : 1 Benzene : 2

p01 = 50.71 mm Hg p02 = 32.06 mm Hg

w1 = 80 g w2 = 100 g

y1 = ?

Molar mass of Benzene(C6H6)

M1 = 6(12) + 6(1)

= 78 g.mol–1

Molar mass of toluene(C6H5CH3)

M2 = 7(12) + 8(1)

= 92 g.mol–1

Moles of benzene (n1) = = = 1.02

Moles of toluene (n2) = = = 1.087

Mole-fraction of benzene (x1) =

= 0.484

Mole-fraction of toluene (x2) = 1 – 0.484

= 0.515

According to Raoult's Law

pTotal = . x1 + . x2

= (50.71)(0.484) + (32.06)(0.515)

= 24.54 + 16.51

= 41.05 mm Hg

Mole-fraction of benzene in vapour phase,

p1 = y1 . pTotal

\ . x1 = y1 . pTotal

\ = y1

\ y1 = y1 = 0.6

//M3//QN39//SUB//DL0//EQ

The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298K if the Henry's law constants for oxygen and nitrogen at 298K are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.

//X

Percentage of O2 = 20%

Percentage of N2 = 79%

Total Pressure = 10 atm

Partial pressure of Oxygen

= × 10 × 760 mm Hg(∵ 1 atm = 760 mm Hg)

= 1520 mm Hg

Partial pressure of Nitrogen

= × 10 × 760 mm Hg

= 6004 mm Hg

Now according to Henry's Law:

For Oxygen:

= KH .

\ =

=

= 460.6 × 10–7

= 4.6 × 10–5

For Nitrogen:

= KH .

\ =

=

= 922.1 × 10–7

= 9.22 × 10–5

//M2//QN40//SUB//DL0//EQ

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

//X

i = 2.47 V = 2.5 litre

p = 0.75 atm

T = 27 + 273 = 300K

R = 0.082 atm.L.mol–1.K–1.

w = ?

M = Ca + 2Cl

= 40 + 2(35.5)

= 111 g.mol–1

p = i .

\ W =

= = 3.42 g

//M2//QN41//SUB//DL0//EQ

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated.

//X

K2SO4 → 2K+ + , T = 25 + 273

= 298 K

Total number of ions(i) = 3, W = 25 mg

= 0.025 g

R = 0.082 atm.L.K–1.mol–1 V = 2 L

M = 174 g.mol–1

p = i .

=

= 0.005266

p = 5.27 × 10–3 atm

Class 12 Chemistry (Part 1) 004