Solutions

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

Given:

2% non-volatile solute in an aqueous solution means that if the solution is of mass 100 g then the mass of solute in the solution is,

Mass of solute = [2 * 100]/100

= 2 g

∴ Mass of Solvent = 100-2

= 98 g

Vapour pressure of solution at normal boiling point, P₁ = 1.004 bar

Vapour pressure of pure water at normal boiling point, P₁° = 1 atm = 1.013 bar

To find: Molar mass of solute

Formula:

From Raoult's Law, we have:

Now we know that number of moles of solute is given by the following relationship,

⇒ Number of moles = Mass of solute / Molecular mass of solute

Using the above relationship the equation [1] can be modified as follows:

Where

W₁ = mass of solvent

W₂ = mass of solute

M₁ = molecular mass of solvent

M₂ = molecular mass of solute

Solution:

(1.013 - 1.004) / 1.013 = 2 X 18 / 98 X M₂

0.009 / 1.013 = 36 / 98M₂

M₂ = 36 X 1.013 / 98 X 0.009

M₂ = 41.3469 g

∴ M₂ ≈ 41.35 g

Therefore the molar mass of the solute is 41.35 g.

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:

The solubility of a gas in water depends on following three parameters:

1. Nature of gas
2. Temperature
3. Pressure

The solubility decreases with increase in temperature. Temperature and pressure follow inverse proportionality. So solubility increases with increase with pressure. A quantitative relation between pressure and solubility of a gas in a solvent was given by W. Henry [1803]. This relationship is known as Henry's law.

Statement:

Henry's law can be expressed as follows.

At constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

Mathematically,

Solubility? Pressure of the gas

Some of the important applications of Henry's law are as follows.

(i) Since the solubility of a gas in water increases with pressure, soft drink bottles are sealed under high pressure to accommodate more CO₂ in the soft drink making the drink fizzier.

(ii) If deep sea divers [scuba divers] use air for respiration, they develop a medical condition known as bends with involves the blocking of capillaries. This is because air is mainly a mixture of oxygen and According to Henry's Law, the solubility of gases increases with increase in pressure. When diver breathes air under high pressure in water, nitrogen dissolves in his blood. When the diver comes towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles in the blood. This is quite painful and dangerous to life. In order to avoid bends and toxic effects of dissolved nitrogen in blood, the tanks used by sea divers are filled with air diluted with helium [ 11.7% H_e + 56.2% N₂ + 32.1% O₂ ].

(iii) The pressure of oxygen in air decreases in going up the mountains. At very high altitudes the partial pressure of oxygen in air is much less than that at the ground Therefore, people living at high altitude or climbers have low concentrations of oxygen in the blood and tissues. This leads to weakness and loss in the clarity of thinking. These symptoms create a condition known as anoxia.

(iv) Henry's law is also related to biology as it explains the supply of inhaled oxygen to tissues. When air is inhaled, it combines with haemoglobin [oxygen career of RBCs] of the blood in lungs to form oxyhaemoglobin because in lungs the partial pressure of oxygen is high. Partial pressure of oxygen is low in tissues. Hence, oxygen is released from oxyhaemoglobin and is utilised by the cells to carry out their tasks.

The lower members of alcohols are completely miscible [highly soluble] with water but the solubility decreases with increase in the molecular weight. The lower members of the alcohol group have the capability to form intermolecular hydrogen bonding with water molecules as alcohols are polar molecules in nature.

Alkyl groups are hydrophobic [prevents formation of hydrogen bonds with water] in nature. In lower alcohols, the alkyl group is small and the –OH group of alcohol is effective in making hydrogen bonds with water.

But with the increase in the size of alkyl group, the hydrophobic [water hating] nature of alkyl group dominates over the hydrophilic [water liking] nature of –OH group making the molecule less soluble in water. So solubility of alcohols decreases with increase in its molecular mass.

Since molecular interaction is weaker between higher alcohols and water, as a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily get liberated off. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.

Given:

Level of contamination = 15 ppm [by mass]

To find: Mass Percentage and Molality

Formula:

Molality = Number of moles of solute / Mass of solvent in kg

Mass Percentage of Solute  = Mass of solute  / Mass of solution X 100

Solution:

Calculation of Mass Percentage:

15 ppm means 15 parts of Chloroform in 10⁶ parts of drinking water

⇒ Mass Percentage = Mass of choloroform / Total mass X 100

= 15 / 10⁶ X 100

= 1.5 * 10^-3

Calculation of Molality:

⇒ Molecular Mass of Chloroform, CHCl₃ = [12] + [1] + [35.5 * 3]

= 119.5 g

⇒ Number of Moles of Chloroform = [15 / 119.5]

= 0.1255 moles

Molality = Number of moles of solute / Mass of solvent in kg

= 0.1255 / 10⁶ X 1000

= 1.255 * 10^-4

Therefore the Mass Percentage is = 1.5 * 10^-3 and the Molality of the solution is = 1.255 * 10^-4 m.

Given:

Mass of ethylene glycol (C₂H₆O₂) = 222.6 g

Mass of water = 200 g

Density, d = 1.072 g/ml

To find: Molality and Molarity of solution

Formula:

Molality = Number of moles of solute/ Mass of solvent in kg

Molarity, M_o = number of moles of solute/ volume of solution in litres

Density, d = Mass (M) / volume (V)

Calculation of Molality:

⇒ Molecular Mass of ethylene glycol (C₂H₆O₂) = [12 * 2] + [6 * 1] + [16 * 2]

= 24 + 6 + 32

= 62 g

⇒ Number of moles of ethylene glycol (C₂H₆O₂) = [222.6/62]

= 3.59 moles

⇒ Mass of water = 200 g

⇒ Molality = numebr of moles of solute/ Mass of solvent in kg

= 3.59 / 200 X 1000

= 17.95 m

Calculation of Molarity:

⇒ Total mass of solution = [222.6 + 200]

= 422.6 g

From density formula we can find out the volume required.

⇒ Volume of solution, V = [422.6 /1.072]

= 394.216 ml

⇒ Molarity, M_o = number of moles of solute/ volume of solution in litres

⇒ Molarity, M_o =3.59 / 394.216 X 1000

⇒ Molarity = 9.1067

⇒ Molarity ≈ 9.11 M

Therefore the Molality and Molarity of the solution is as follows:

Molality = 17.95 m

Molarity = 9.11 M