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?
The boiling point of substance can be defined as, it is the temperature at which the vapour pressure of the substance equal to atmospheric pressure. According to Raoult's law, the vapour pressure of a solvent decreases in the presence of a non-volatile solute. Thus, the vapour pressure of a solution containing a non-volatile solute requires a high temperature to become equal to the atmospheric pressure. That is why the boiling point of a solution containing a non-volatile solute increases. Thus, the solution containing a non-volatile solute has a higher boiling point than a pure solvent. Example of non-volatile solute sugar, NaCl. etc.
Elevation in boiling point is a colligative property, as it depends on the number of solute particles present in a solution.
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)
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