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

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

//X

Refer Que. 1 (Page No.20)

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

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

//X

Refer Que. 2 (Page No.20)

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

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

//X

(i) Refer Que. 8 (Page No.21)
(ii) Refer Que. 10 (Page No.22)
(iii) Refer Que. 9 (Page No.22)
(iv) Refer Que. 4 (Page No.20)

//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 HNO 3 = 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 H 2 O = 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 Na 2 CO 3 = x g
∴ Mass of NaHCO 3 = (1 – x) g
Molar mass of Na 2 CO 3 = 106 g.mol–1
Molar mass of NaHCO 3 = 84 g.mol–1
Na 2 CO 3 and NaHCO 3 both are equimolar
∴ Moles of Na 2 CO 3 = Moles of NaHCO 3
∴ =
∴ 84 x = 106 – 106 x
∴ 84 x + 106 x = 106
∴ 190 x = 106
∴ x = 0.5578 g
∴ Mass of Na 2 CO 3 = x = 0.5578 g
∴ Mass of NaHCO 3 = 1 – x
= 1 – 0.5578
= 0.4422 g
Let us find out mass of HCl which react with Na 2 CO 3 and NaHCO 3
Na 2 CO 3 + 2 HCl → 2 NaCl + H 2 O + CO 2
106 g = 2 × 36.5
= 73 g
If 106 g Na 2 CO 3 react with 73 g of HCl
Then 0.5578 g Na 2 CO 3 react with:
Mass of HCl =
= 0.384 g
NaHCO 3 + HCl → NaCl + CO 2 + H 2 O
84 g 36.5 g
If 84 g NaHCO 3 react with 36.5 g of HCl
Then 0.4422 g NaHCO 3 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
m olality =
= 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

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.

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

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

//X

Refer Que. 12 (Page No.23)

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

State Henry's law and mention some important applications.

//X

Refer Que. 14 (Page No.23)

//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 = K H .
\ P = K H .
Moles of Solute (C 2 H 6) << Moles of Solvent
\ 1 = K H .
\ K H =
Moles of ethane =
= 0.166 × 10–2
= 1.66 × 10–3
P = K H . 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 D mixH related to positive and negative deviations from Raoult's law?

//X

Refer Que. 19 (Page No.28)

//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
w 2 = 2 g M 2 = ?
w 1 = 98 g M 1 = 18 g.Mol–1
p 1 = 1.004 bar = 1.013 bar
=
=
\ =
\ M 2 =
= 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
w 1 = 26.0 g w 2 = 35 g
p Total = ?
Molar mass of heptane (M 1) = 7(C) + 16(H)
= 7(12) + 16(1)
= 84 + 16 = 100 g.mol–1
Molar mass of octane (M 2) = 8(C) + 18(H)
= 8(12) + 18(1) = 96 + 18
= 114 g.mol–1
Moles of heptane (n 1) =
=
= 0.26 Mol
Moles of octane (n 2) =
=
= 0.31 Mol
Mole-fraction of heptane (x 1) =
= 0.456
Mole-fraction of octane (x 2) = 1 – 0.456
= 0.544
According to Raoult's Law
p Total = p 1 + p 2
= . x 1 + . x 2
= (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
p 1 = ?
1 molal solution means that 1 mole of solute in 1000 g of solvent
n 2 = 1, n 1 = = 55.55
=
=
\ 12.3 – p 1 =
\ 12.3 – p 1 = 0.2175
\ p 1 = 12.3 – 0.2175
\ p 1 = 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
w 2 = ? M 2 = 40 g.mol–1
w 1 = 114 g M 1 = (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

w 2 = 30 g w 1 = 90 g
M 2 = ? p 1 = 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
w 2 = 5 g w 1 = 95 g
M 2 = 342 g/mol T f = 271K, = 273.15K
D T f = – T f
= 273.15 – 271
= 2.15K
D T f = K f .
\ K f =
=
K f = 13.97 K.kg.mol–1
Glucose: 5% W/W
w 2 = 5g w 1 = 95 g
M 2 = 180 g.mol–1 D T f = ?
D T f = K f .
= 13.97 ×
D T f = 4.08K
\ D T f = – T f
\ 4.08 = 273.15 – T f
\ T f = 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

M 2 = K f .
For AB 2: w 2 = 1 g w 1 = 20 g
= ? D T f = 2.3K
K f = 5.1 K.kg.mol–1
M AB 2 = = 110.87 g.mol–1
For AB 4: w 2 = 1 g w 1 = 20 g
D T f = 1.3K M AB 4 = ?
M AB 4 =
= 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, w 2 = 36 g, M 2 = 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

(i) Van der Waal's forces of attraction.
(ii) Van der Waal's forces of attraction.
(iii) Ion-dipole interaction.
(iv) Dipole-dipole interaction.
(v) Dipole-dipole interaction.

//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

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 < CH 3 CN < CH 3 OH < KCl
Therefore, the order of increasing solubility is:
KCl < CH 3 OH < CH 3 CN < Cyclohexane

//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

(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 (C 5 H 11 OH) has polar –OH group, but is also contains a very bulky non-polar -
–C 5 H 11 group. Thus, pentanol is partially soluble in water.

//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
K SP = [Cu+2] . [S–2]
\ 6 × 10–16 = S . S
\ 6 × 10–16 = S 2
\ 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

Mass of benzoic acid = ?
Volume of solution = 250 mL
= 0.25 L
Molarity = 0.15 M
Molar mass of benzoic acid = 122 g.mol–1
M =
\ Mass of benzoic acid = 0.15 × 122 × 0.25
= 4.575 g

//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

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.

//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

w 2 = 10 g, w 1 = 250 g
M 2 = 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 K f = 1.86 K.kg.mol–1
D T f = ?
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
D T f = i . K f . 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

w 2 = 19.5 g w 1 = 500 g
M 2 = 78 g.mol–1 D T f = 1.0°C
K f = 1.86 K.kg.mol–1
i = ? K a = ?
D T f = i . K f .
\ i =
=
\ i = 1.07526
∝ =
\ ∝ =
\ ∝ = 0.07526
Molality =
=
= 0.5131
K a =
=
=
= 0.00313
K a = 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

w 2 = 25 g w 1 = 450 g
M 2 = 180 g.mol–1 M 1 = 18 g.mol–1
= 17.535 p 1 = ?
n 1 = = = 25
n 2 = = = 0.14
\ =
\ =
\ 17.535 – p 1 =
\ 17.535 – p 1 = 0.0976
\ p 1 = 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

K H = 4.27 × 10 5 mm Hg
P = 760 mm Hg
Solubility (X) = ?
According to Henry's Law
P = K H . 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
W A = 100 g W B = 1000 g
M A = 140 g.Mol–1 M B = 180 g.Mol–1
p 0 A = (?) p 0 B = 500 torr
p A = (?)
p Total = 475 torr
Moles of liquid–A (n A) = = = 0.714
Moles of liquid–B (n B) = = = 5.55
Mole-fraction of liquid : A (x A) =
= 0.114
Mole-fraction of liquid : B (x B) = 1 – 0.114
= 0.886
p Total = p 0 A . x A + p 0 B . x B
475 = p 0 A . (0.114) + (500)(0.886)
\ 475 = p 0 A . (0.114) + 443
\ 475 – 443 = p 0 A . (0.114)
\ 32 = p 0 A . (0.114)
\ p 0 A = 280.7 torr
Vapour pressure of pure liquid A is 280.7 torr
Vapour pressure of liquid A gaseous in solution
p A = p 0 A . x A
= (280.7) . (0.114)
p A = 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 p total, p chloroform, and p acetone as a function of x acetone. The experimental data observed for different compositions of mixture is: 100 X x acetone 0 11.8 23.4 36.0 50.8 58.2 64.5 72.1 p acetone/mm Hg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1 p chloroform/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 acetone | 0 | 0.118 | 0.234 | 0.360 | 0.508 | 0.582 | 0.645 | 0.721
p acetone | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 527.1
p chloroform | 632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7
p Total | 632.8 | 603.0 | 579.5 | 562.1 | 580.4 | 599.5 | 615.3 | 641.8

//X

In the graph p total 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.

OR
Benzene and toluene from ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300K are 50.7 mm Hg and

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.

(Atomic mass : C = 12 u, H = 1 u).

//X

Ans. Benzene : 1 Benzene : 2
p 0 1 = 50.71 mm Hg p 0 2 = 32.06 mm Hg
w 1 = 80 g w 2 = 100 g
y 1 = ?
Molar mass of Benzene(C 6 H 6)
M 1 = 6(12) + 6(1)
= 78 g.mol–1
Molar mass of toluene(C 6 H 5 CH 3)
M 2 = 7(12) + 8(1)
= 92 g.mol–1
Moles of benzene (n 1) = = = 1.02
Moles of toluene (n 2) = = = 1.087
Mole-fraction of benzene (x 1) =
= 0.484
Mole-fraction of toluene (x 2) = 1 – 0.484
= 0.515
According to Raoult's Law
p Total = . x 1 + . x 2
= (50.71)(0.484) + (32.06)(0.515)
= 24.54 + 16.51
= 41.05 mm Hg
Mole-fraction of benzene in vapour phase,
p 1 = y 1 . p Total
\ . x 1 = y 1 . p Total
\ = y 1
\ y 1 = y 1 = 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 N 2 = 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:
= K H .
\ =
=
= 460.6 × 10–7
= 4.6 × 10–5
For Nitrogen:
= K H .
\ =
=
= 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