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Solutions

Short Answer Type

211. Give one example each of miscible liquid pairs showing positive and negative deviations from Raoult’s law. Give one reason for such deviations.

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212. State Raoult’s law for solutions where only solvent is volatile. Derive a mathematical exression for this law.

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213.

State with a suitable diagram and appropriate examples why some non-ideal solutions. Show positive deviation from ideal behaviour.

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214. What is meant by abnormal mass of solute? Discuss the factors which bring abnormality in the experimentally determined molecular masses of solutes using colligative properties.

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215. State the type of non-ideality exhibited by a solution of cyclohexane and ethanol or a solution of acetone and chloroform (only one case). Give reason for your answer.

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216. Miscible liquid pairs often show negative or positive deviation from Raoult’s law. What is the reason for such deviations? Give one example of each type of liquid pairs.

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Long Answer Type

217. Define vapour pressure of a liquid. What happens to the vapour pressure when (a) a volatile solute dissolves in the liquid and (b) non-volatile solute dissolvedin it?

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218. Show graphically how the vapour pressure of a solvent and a solution in it of a non-volatile solute change with temperature. Show on this graph the boiling points of the solvent and the solution. Which is higher and why?

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219. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₄(OH)₂) and 200 g of water. Calculate the molality of the solution. If the density of this solution be 1.072 g ml^–1, what will be the molarity of the solution?

Answer:

Molecular mass of glycol C₂H₄(OH)₂ =62

Number of moles of C₂H₄(OH)₂
= mass of glycol / molar mass of glycol

= 222.6/62=3.59 moles

Molality of solution =
number of moles of solute / mass of solvent in Kg

Mass of solvent = 200g = 200/1000 Kg = 0.2 kg

Plug the values we get

Molality = 3.59/0.2

Molality = 17.95

Formula of molarity of solution = number of moles of solute / volume of solution in Kg

Formula of volume = mass / density

Volume = 422.6g/(1.072 g/ml)

Volume = 394.21 ml

Convert in liter

Volume in liter = 394.21 ml /1000 liter =0.394 liter

Molarity =9.1M

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220. At 300 K, 36 g of glucose (C₆H₁₂O₆) present per litre in its aqueous solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of another solution of glucose is 1.52 bar at the same temperature, what would be its concentration?

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