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What type of azeotrope is formed by positive deviation from Raoult’s law? - Chemistry

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प्रश्न

What type of azeotrope is formed by positive deviation from Raoult’s law?

अति संक्षिप्त उत्तर
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उत्तर

Azeotropes exhibiting negative deviation from Raoult’s law form maximum boiling azeotropes at a specific composition.

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2014-2015 (March) Delhi Set 1

संबंधित प्रश्‍न

What type of deviation is shown by a mixture of ethanol and acetone? Give reason.


A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

  1. molar mass of the solute.
  2. vapour pressure of water at 298 K.

What is meant by negative deviation from Raoult's law? Give an example. What is the sign of ∆mixH for negative deviation?


For the reaction :

\[\ce{2NO_{(g)} ⇌ N2_{(g)} + O2_{(g)}}\];

ΔH = -heat

K= 2.5 × 10at 298K

What will happen to the concentration of Nif :

(1) Temperature is decreased to 273 K.

(2) The pressure is reduced


Match the following:

(i) Colligative property (a) Polysaccharide
(ii) Nicol prism (b) Osmotic pressure
(iii) Activation energy (c) Aldol condensation
(iv) Starch (d) Polarimeter
(v) Acetaldehyde (e) Arrhenius equation

An aqueous solution of hydrochloric acid:


Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law?


On the basis of information given below mark the correct option.

(A) In bromoethane and chloroethane mixture intermolecular interactions of A–A and B–B type are nearly same as A–B type interactions.

(B) In ethanol and acetone mixture A–A or B–B type intermolecular interactions are stronger than A–B type interactions.

(C) In chloroform and acetone mixture A–A or B–B type intermolecular interactions are weaker than A–B type interactions.


On the basis of information given below mark the correct option.
On adding acetone to methanol some of the hydrogen bonds between methanol molecules break.


Using Raoult’s law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.

\[\ce{NaCl(s) and H2O(l)}\]


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