Subject

Chemistry

Class

CBSE Class 12

Pre Boards

Practice to excel and get familiar with the paper pattern and the type of questions. Check you answers with answer keys provided.

Sample Papers

Download the PDF Sample Papers Free for off line practice and view the Solutions online.
Advertisement

 Multiple Choice QuestionsLong Answer Type

31.

(a) Complete the following chemical equation

(i) Cu + HNO3 (dilute) ---> 

(ii) XeF4 + O2F2 -->

(b) Explain the following observation:

(i) Phosphorus has a greater tendency for catenation than nitrogen.

(ii) Oxygen is a gas but sulphur a solid.

(iii) The halogens are coloured. Why?

998 Views

32.

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)

1146 Views

Advertisement

33.

(a) Explain the following:

(i) Henry’s law about the dissolution of a gas in a liquid.

(ii) Boiling point elevation constant for a solvent.

(b) A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass if glycerol was dissolved to make this solution? (Kb for water = 0.512 K kg mol-1)


(a)

(i) Henry’s law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas, then Henry’s law can be expressed as:

p = KHx

Where,

KH is Henry’s law constant

(ii) The boiling point elevation constant or molal elevation constant is a constant quantity for a solute which is related to molar mass and elevation in boiling point by the following relation.

 

Where, Kb is the boiling point elevation constant

MB is the molar mass of the solute

WB is the weight of the solute

WA is the weight of the solvent

Tb is the elevation in boiling point

(b) WB =?

WA = 500g

Kb = 0.512 Kkg mol-1

Tb = 100.42°C - 100°C

= 0.42°C

MB = 3 x 12 + 8 x 1 + 3 x 16

= 36 + 8 + 48 = 92

 

1757 Views

Advertisement
34.

(a) Write a suitable chemical equation to complete each of the following transformations:

(i) Butan-1-ol to butanoic acid

(ii) 4-methylacetophenone to benzene-1, 4-dicarboxylic acid

(b) An organic compound with molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.

1378 Views

Advertisement
35.

(a) Give chemical tests to distinguish between

(i) Propanol and propanone

(ii) Benzaldehyde and acetophenone

(b) Arrange the following compounds in an increasing order of their property as indicated:

(i) Acetaldeyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)

(ii) Benzoic acid, 3, 4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength).

(iii) CH3CH2CH (Br) COOH, CH3CH (Br) CH2COOH, (CH3)2CHCOOH (acid strength)

1283 Views

Advertisement