Solutions

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

Given:

Molarity of HCl, = 0.1 M

Mass of Mixture = 1g

To find: Volume of HCl to react completely with mixture

Formula:

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

Solution:

Calculation of Amount of each component in mixture:

⇒ Let the amount of Na₂CO₃ be X g

⇒ And Let amount of NaHCO₃ be [1-X] g

⇒ Molecular Weight of Na₂CO₃ = [23 * 2] + [12] + [3 * 16]

= 106 g

⇒ Molecular Weight of NaHCO₃ = [23] + [1] + [12] + [3 * 16]

= 84 g

⇒ Number of moles of NaHCO₃ = 1-x / 84

⇒ Number of moles of Na₂CO₃ = x / 106

Now it is given in the question that the mixture is equimolar, so

⇒ Number moles of Na₂CO₃ = Number of moles of NaHCO₃

1-x/84 = x/106

106 – 106X = 84X

190X = 106

x= 106/190

X = 0.5579

≈ 0.558

Amount of Na₂CO₃ = 0.558 g

⇒ Amount of NaHCO₃ = 1- 0.558 g = 0.442 g

⇒ Number moles of Na₂CO₃ = [0.558/106]

= 0.00526 moles

⇒ Number of moles of NaHCO₃ = 0.00526 moles …. [Mixture is equimolar]

Calculation of volume of HCl required

⇒ In the question it is given that HCl reacts with the mixture.

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

NaHCO₃ + HCl → NaCl + H₂O + CO₂

From the above chemical equations it can be found that 2 moles of HCl is required to react with one mole of Na₂CO₃ and 1 moles of HCl is required to react with one mole of NaHCO₃.

So total number of moles of HCl required to completely react with mixture is as follows:

⇒ Total Number of Moles of HCl = [2 * 0.00526] + 0.00526

= 0.01578 moles

From Molarity Formula we have,

Molarity, M_o = number of moles of solute/ volume in litres (V)

⇒ Volume in litres, V = [0.01578/0.1]

= 0.1578 L

⇒ Volume in ml, V = 0.1578 * 1000

= 157.8 ml

Therefore the volume of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both is 157.8 ml.

Given:

Concentration of Nitric Acid, HNO₃ = 68%

Density of solution, d = 1.504 g/ml

To find: Molarity, M_o

Formula:

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

Molarity, M_o  = Number of moles of solute/ Volume of solution in litres

Solution:

68% of Nitric acid by mass in aqueous solution means that 68g [68 * 100]/100] of Nitric acid present in 100g of solution.

⇒ Molecular mass of Nitric Acid, HNO₃ = [1 * 1] + [1 * 14] + [16 * 3]

= 63g

⇒ Number of moles of Nitric Acid = [68/63]

= 1.079 moles

⇒ Given Density, d = 1.504 g/ml

⇒ Volume, v = [100/1.504]

= 66.489 ml

⇒ Molarity, M_o = [1.079/66.489] * 1000

= 16.23 M

Therefore the molarity of the sample is 16.24 M.