State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations.

Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. Consider a binary solution of two components A and B (both liquids). Let p total be the total vapour pressure of the solution due to both the components and let p A and p B be the partial pressures of component A and B , respectively. Then, p A = x A p A ° p B = x B p B ° The total vapour pressure of the solution of liquids A and B is then, by Dalton’s law of partial pressures, the sum of p A and p B . p Total = p A + p B Substituting for p A and p B , we get p Total = x A p A ° + x B p B ° The characteristics of the solution which obeys Raoult’s law at all concentrations are as follows: (i) In a binary solution of components, A and B , the enthalpy of mixing ΔH mix is zero, that is, in preparation of an ideal solution no thermal change is observed. (ii) In an ideal solution, the volume of mixing ( ΔV mix ) is also zero, that is, the final volume of the solution is equal to the sum of volumes of components being mixed.

Visha took two aqueous solutions — one containing 7·5 g of urea (Molar mass = 60 g/mol) and the other containing 42·75 g of substance Z in 100 g of water, respectively. It was observed that both the solutions froze at the same temperature. Calculate the molar mass of Z.

Given, 0.75 g of urea Molar mass = 60 g/mol 42.75 g of the substance of z 100 g of water It is also mentioned that both the solutions froze at the same temperature. ∆ T f × W . urea M . urea × 1000 w . solvent = kf × w . z M . z × 1000 W . solvent 7 . 5 60 × 1000 100 = 42 . 75 Mz × 1000 100 Mz = 42 . 75 × 60 7 . 50 = 342 g / mol .

Identify which liquid will have a higher vapour pressure at 90 ℃ if the boiling points of two liquids A and B are 140 ℃ and 180 ° C , respectively.

Liquid A has lower boiling point than that of liquid B and due to its low boiling point, liquid A has weak intermolecular forces than liquid B. Thus, the tendency of liquid A to vaporize will be greater than liquid B and its vapour pressure will be higher at 90 ° C .

Calculate the mass of ascorbic acid Molar Mass = 176 g mol - 1 to be dissolved in 75 g of acetic acid, to lower its freezing point by 1 . 5 ° C . K f = 3 . 9 K kg mol - 1

Given mass of acetic acid is w 1 = 75 g Molar mass of ascorbic acid is M 2 = 176 g mol - 1 Freezing point is defined as the temperature at which liquid state of a substance changes to solid state. Given that, lowering of freezing point is ∆ T f = 1 . 5 K We know that, ∆ T b = k b × w 2 × 1000 M 2 × w 1 w 2 = M 2 × w 1 × ∆ T b k b × 1000 w 2 = 1 . 5 × 176 × 75 3 . 9 × 1000 = 5 . 08 g Mass of ascorbic acid is w 2 = 5 . 08 g

Give reasons for the following: Elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.

KCl being an electrolyte dissociates almost completely to give K + and Cl − ions whereas glucose being non-electrolyte does not dissociate. Hence, the number of particles in 1 M KCl solution is nearly double that in the 1 M sugar solution. Elevation of boiling point, being a colligative property (these properties depend on the number of solute particles) is, therefore, nearly twice for KCl solution than for glucose solution of same molarity.

Give reasons for the following: Aquatic animals are more comfortable in cold water than in warm water.

At a given pressure, the solubility of oxygen in water increases with a decrease in temperature. Therefore, the concentration of oxygen in the sea is more in cold water and thus the presence of more oxygen at lower temperatures makes the aquatic species more comfortable in cold water.

Give reasons for the following: Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.

Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers because of the following two reasons: (1) The values given by the other colligative properties (elevation in the boiling point, the depression in the freezing point and relative lowering of vapour pressure) are so low that they cannot be measured accurately. (2) Measurement of osmotic pressure is carried out at room temperature which does not require heating. Heating may change the nature of the polymer.

Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol - 1 ) in 250 g of water ( K f = 1 . 86 K kg mol - 1 ).

Let us first calculate molality of the solution. Mathematically, m = W solute M solute × 1000 W solvent ( in g ) .............(i) On substituting the given values in the equation (i), m = 60 180 × 1000 250 = 1 . 33 mol kg - 1 ∆ T f = K f m ⇒ ∆ T f = 1 . 86 × 1 . 33 ≈ 2 . 48 K Depression in freezing point is the difference between freezing points of solution and of pure solvent respectively. i.e., ∆ T f = T ∘ - T 1 ∴ 273 - T 1 = 2 . 48 Or, T 1 = 270 . 52 K The freezing point of the solution is 270 . 52 K .

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For a 5% solution of urea (Molar mass = 60 g/mol), calculate the osmotic pressure at 300 K. [ $R = 0 \cdot 0821 L atm K^{- 1} {mol}^{- 1}$ ]

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