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 Multiple Choice QuestionsShort Answer Type

251.

A 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)

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252.

Define the following terms:

 (i) Mole fraction

(ii) Isotonic solutions

(iii) Van’t Hoff factor

 (iv) Ideal solution

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253.

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)

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 Multiple Choice QuestionsLong Answer Type

254.

(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

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255.

(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)

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 Multiple Choice QuestionsShort Answer Type

256.

Why does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property?

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257.

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)

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 Multiple Choice QuestionsLong Answer Type

258.

(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

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259.

(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)

(i) Ideal Solution:

Solutions which obey Raoult’s law over the entire range of concentrations are known as the 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

(ii) Azeotropes

Binary mixtures which have the same composition in liquid and vapour phase, and have constant boiling points are known as azeotropes. It is not possible to separate its components by fractional distillation .There are two types of azeotropes:

Minimum boiling azeotrope, example: Ethanol-water mixture containing ethanol approximately 95% by volume.

Maximum boiling azeotrope, example: Nitric acid-water mixture containing 68% nitric acid and 32% water by mass.

(iii) Osmotic Pressure

The process of flow of solvent molecules from pure solvent to a solution or from a solution of lower concentration to a solution of higher concentration through a semi-permeable membrane is called osmosis. The pressure required to prevent the flow of solvent due to osmosis is called osmotic pressure (Ï€) of the solution.

Osmotic pressure is directly proportional to the molarity C of the solution at a given temperature T.



 

Where,

n2 = Number of moles of solute

V= Volume of the solution in litres

 

(b) Assume that 100 g of solution contains 10 g of glucose and 90 g of water as our glucose solution is 10% by weight

 

Where w2 = weight is solute in 'g'

w1 = weight of solvent in 'g'

M2 = Molecular mass of solute.

So,

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260.

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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