Class 12th

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

16.1 

Sleeping pills belong to tranquilizers class of drugs. Sleeping pills induce artificial sleep in people suffering from insomnia by depressing the activities of the central nervous system [brain and spinal cord].

When such drugs are taken for a prolonged time, such drugs can become addictive and the person can become addicted to such drugs and the aftereffects of such drugs are drowsiness, hallucinations, slow heart rate etc. and in extreme cases, the person can enter a state of coma and death can also take place.

Therefore it is highly recommended that sleeping pills should be administered to people suffering from insomnia but under the strict consultation of a doctor.

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.