Solutions

let the molar masses of AB₂ and AB₄ be x and y respectively.

Molar mass of benzene (C₆H₆) = 12 * 6 + 1 * 6 = 78 g/mol

Moles of benzene = mass/molar mass = 20/78

n = 0.256mol

⇒ ΔT_f = 2.3 K

K_f = 5.1K kg mol^-1

For AB₂

Applying the formula: ΔT_f = K_f * M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ 2.3 = 5.1 * M₁

⇒ M₁ = 0.451mol/kg

For AB₄

Applying the formula: ΔT_f = K_f * M

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ 1.3 = 5.1 * M₂

⇒ M₂ = 0.255mol/kg

M₁ = moles of solute/mass of solvent (in kg)

M₁ = 1/x / 0.02 = 1 / 0.02x = 0.451

⇒ X = 110.86g

M₂ = moles of solute/mass of solvent (in kg)

M₂ = 1/y / 0.02 = 1 / 0.02y = 0.255

⇒ y = 196.1g

atomic mass of A be a and that of B be b g respectively.

So, AB₂ : a + 2b = 110.86

AB_4: a + 4b = 196.1

⇒ a = 25.59g

⇒ b = 42.64g

5% solution means 5g of cane sugar is present in 100g of solution

Freezing point of solution = 271k

Freezing point of pure water = 273.15k

Molar mass of cane sugar (C₁₂H₂₂O₁₁) = 12 * 12 + 1 * 22 + 16 * 11 = 342g

Moles of cane sugar = mass/molar mass = 5/342

⇒ n = 0.0146mol

Molality of solution = moles of solute/mass of solvent (in kg)

⇒ M = 0.0146/0.095

⇒ Molality = 0.154M

Depression in freezing point = ΔT_f = 273.15-271 = 2.15k

Applying the formula: ΔT_f = K_f * M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ K_f = 2.15/0.154

⇒ K_f = 13.96k kg mol^-1

Second condition: mass of glucose = 5g

Molar mass of glucose (C₆H₁₂O₆) = 12 * 6 + 1 * 12 + 16 * 6 = 180g

Moles of glucose = mass/molar mass

⇒ n = 5/180moles

⇒ n = 0.0278mol

Molality of solution = moles of solute/mass of solvent (in kg)

⇒ molality = 0.0278/0.095

⇒ M = 0.2926M

Applying the formula: ΔT_f = K_f * M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ ΔT_f = 13.96 * 0.2926

⇒ ΔT_f = 4.08k

Freezing point of solution = 273.15 + 4.08 = 277.234k

Given: mass of solute = 30g

Let the molar mass of solute be x g and vapour pressure of pure water at 298k be P₁ ^?

Mass of water(solvent) = 90g

Molar mass of water = H₂O = 1 * 2 + 16 = 18g

Moles of water = mass of water/molar mass

⇒ n = 90/18 moles

⇒ n = 5moles

Molar fraction of solute,

x₂ = moles of solute / moles of solute + moles of octane

x₂ = (30/x) / (30/x) + 5

x₂ = 30 / 30+5x

Vapour pressure of solution (p₁) = 2.8kpa

Applying the formula:

According to second condition when we add 18g of water to solution vapour pressure becomes 2.9kpa

Moles of water = mass/molar mass

⇒ n = 90 + 18/18

⇒ n = 6moles

Molar fraction of solute,

x₂ = moles of solute / moles of solute + moles of octane

x₂ = (30/x) / (30/x) + 6

x₂ = 30 / 30+6x

Applying the formula:

Solving 1 and 2:

Dividing (2) by (1) we get

Substituting value of x in 1 we get

(i) molar mass of the solute = 78g

(ii) vapour pressure of water at 298 K = 537kpa

Given:

Molar mass of non-volatile solute = 40g

Let no. of moles of solute be n.

Mass of octane = 114g

Molar mass of octane (C₈H₁₈) = 12 * 8 + 1 * 18 = 114g/mol

Moles of octane = given mass/molar mass

⇒ n = 114/114 moles

⇒ n = 1 mole

Molar fraction of solute,

x₂ = moles of solute / moles of solute + moles of octane

⇒ x₂ = n/n + 1

Let the vapour pressure of original solvent (without solute) be p₁^?

Accordingly after addition of solute vapour pressure of solution reduces to 80% i.e.

0.8 p₁^? = p₁

Applying the formula:

⇒ n/n + 1 = 0.2

⇒ 0.2n + 0.2 = n

⇒ n = 0.25 moles

Hence, mass of solute is:

moles = given mass/molar mass

⇒ 0.25moles = mass/40g

⇒ mass = 10g.

Given: Temperature = 373k

Vapour pressure of pure heptane (p₁⁰ ) = 105.2 kpa and that of octane (p₂⁰ ) = 46.8 kpa

Mass of heptane = 26 g

Mass of octane = 35 g

Molecular weight of heptane = C₇H₁₆ = 12 * 7 + 1 * 16 = 100 gmol^-1

Molecular weight of octane = C₈H₁₈ = 114 gmol^-1

Moles of heptane, n₁ = given mass /molecular weight = 26/100

⇒ n₁ = 0.26mol

Moles of octane, n₂ = given mass /molecular weight = 35/114

⇒ n₂ = 0.307mol

∴ Partial pressure of heptane, p₁ = χ₁ * p₁⁰

⇒ p₁ = 0.456 * 105.2 = 47.97kpa

∴ Partial pressure of octane, p₁ = χ₂ * p₂⁰

⇒ p₂ = 0.544 * 46.8 = 25.46 kpa

∴ Total pressure exerted by solution = p₁ + p₂

= 47.97 + 25.46

= 73.43 kpa

Raoult's law states that at a given temperature, the vapour pressure of a solution containing non volatile solute is directly proportional the mole fraction of the solvent.

Non ideal solutions show positive and negative deviations from ideal behaviour.

Non ideal solutions showing positive deviations from Raoult's law-

A solution is said to show positive deviation from Raoult's Law when at any composition, its vapour pressure is higher than that given by Raoult's Law.

The positive deviation is shown by those liquid pairs in which the A-B molecular interaction forces are weaker than the corresponding A-A or B-B molecular interaction forces. When the A-B molecular interaction forces are weaker, then molecules of liquid A find it easier to escape as compared to pure solution thereby increasing the vapour pressure of solution.

As a result, each component of solution has a partial vapour pressure greater than expected on the basis of Raoult's law. This is called positive deviations from Raoult's law, that is P_A>P_A^oX_A and P_B>P_B^oX_B

_{? mix}H [Change in Enthalpy] is positive because energy is required to break A-A & B-B attractive forces. Hence it is an endothermic process.

Examples of liquid Pairs:

1. Water + methanol
2. Water + ethanol
3. Acetone + benzene
4. Carbon tetrachloride + benzene

The graph of vapour pressure and mole fraction for liquids showing positive deviation from Raoult's Law is shown below:

From the diagram it is clear that the composition curve of such a solution lies above the composition curve obtained on basis of Raoult's Law

Non ideal solutions showing Negative deviations from Raoult's law –

When the vapour pressure of the solution is lower than that calculated on the basis of Raoult's Law, the solution is said to show negative deviation from Raoult's Law. Even the vapour pressure of individual components show negative deviation from Raoult's Law

The negative deviation is shown by those liquid pairs in which the A-B molecular interaction forces are stronger than the corresponding A-A or B-B molecular interaction forces. When the A-B molecular interaction forces are stronger, then molecules of liquid A find it easier to escape from pure solution as compared to the mixture thereby decreasing the vapour pressure of solution.

Consequently, each component of solution has a partial vapour pressure less than expected on the basis of Raoult's law. This is called negative deviations from Raoult's law, i.e. P_AA^oXA & PB^oXB.

Examples of liquid pairs:

1. Water + hydrochloric acid
2. Water + nitric acid
3. Chloroform + benzene
4. Acetone + aniline

_{? mix}H is negative because energy is released due to increase in attractive forces. Hence it is an exothermic process.

The graph of vapour pressure and mole fraction for liquids showing negative deviation from Raoult's Law is shown below: