Short Answer TypeA 1.00 molar aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling point of 100.180C. Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 kg mol-1)
Define the following terms:
(i) Mole fraction
(ii) Isotonic solutions
(iii) Van’t Hoff factor
(iv) Ideal solution
Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2K. (Kf for water = 1.86 K kg mol-1)
Long Answer Type(a) Differentiate between molarity and molality for a solution. How does a change in temperature influence their values?
(b) Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr2in 200 g of water. (Molar mass of MgBr2 = 184 g) (Kf for water = 1.86 K kg mol-1)
(a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (Kb for water = 0.512 K kg mol-1), (Molar mass of NaCl = 58.44 g)
Short Answer TypeWhy does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property?
Calculate the freezing point of the solution when 31 g of ethylene glycol (C2H6O2) is dissolved in 500 g of water. (Kf for water = 1.86 K kg mol–1)
Long Answer Type(a) State Raoult’s law for a solution containing volatile components.
How does Raoult’s law become a special case of Henry’s law?
(b) 1·00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0·40 K. Find the molar mass of the solute. (Kf for benzene = 5·12 K kg mol-1)
(a) Define the following terms:
(i) Ideal solution
(ii) Azeotrope
(iii) Osmotic pressure
(b) A solution of glucose (C6H12O6) in water is labeled as 10% by weight. What would be the molality of the solution?
(Molar mass of glucose = 180 g mol-1)
a) Define the following terms:
(i) Mole fraction
(ii) Ideal solution
(b) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at - 0.34°C. What is the molar mass of the material? (Kf for water = 1.86 K kg mol-1)
(i) Mole fraction: The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture. Mathematically, it is represented as:
Mole fraction is denoted by ‘x’.
(ii) Ideal Solution:
Solutions which obey Raoult’s law over the entire range of concentrations are known as ideal solution. Along with that for ideal solution:
Enthalpy of mixing of the pure components to form the solution i.e mix H = 0 and volume of mixing, mix V = 0.
An ideal solution will be formed when intermolecular forces of attraction between the molecules of solute (A - A) and those between the molecules of solvent (B -B) are nearly equal to those between solute and solvent molecules (A - B).
For Example: n-Hexane and n-heptane
(b) Given mass of solute = wb = 15.0g
Molar mass of solute = Mb =?
Mass of water = wa = 450 g
Freezing point of water = 0°C = 273 K
Freezing point of solution = - 0.34°K
= (- 0.34 + 273) K
Depression in freezing point = Tf= 273 - (-0.34 + 273)
= 0.34 K
Kf for water = 1.86 K Kg mol-1
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