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8/31/13 NCERT Solutions for Class 12th Chemistry Chapter 2 Solutions

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NCERT Solutions for Class 12th Chemistry Chapter 2 SolutionsSeptember 3, 2012 | Filed under: 12th Class,Book Solutions,Chemistry | Posted by: admin

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Question 2.23:

Suggest the most important type of intermolecular attractive interaction in the following pairs.

(i) n-hexane and n-octane

(ii) I2 and CCl4

(iii) NaClO4 and water

(iv) methanol and acetone

(v) acetonitrile (CH3CN) and acetone (C3H6O).

Answer

(i) Van der Wall’s forces of attraction.

(ii) Van der Wall’s forces of attraction.

(iii) Ion-diople interaction.

(iv) Dipole-dipole interaction.

(v) Dipole-dipole interaction.

Question 2.24:

Based on solute-solvent interactions, arrange the following in order of increasing

solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.

Answer

n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-

octane.

The order of increasing polarity is:

Cyclohexane < CH3CN < CH3OH < KCl

Therefore, the order of increasing solubility is:

KCl < CH3OH < CH3CN < Cyclohexane

Question 2.25:

Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

(i) phenol (ii) toluene (iii) formic acid

(iv) ethylene glycol (v) chloroform (vi) pentanol.

Answer

(i) Phenol (C6H5OH) has the polar group −OH and non-polar group −C6H5. Thus, phenol is partially soluble in water.



(ii) Toluene (C6H5−CH3) has no polar groups. Thus, toluene is insoluble in water.

(iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water.

Thus, formic acid is highly soluble in water.

(iv) Ethylene glycol has polar −OH group and can form H−bond. Thus, it is highly soluble in water.

(v) Chloroform is insoluble in water.

(vi) Pentanol (C5H11OH) has polar −OH group, but it also contains a very bulky nonpolar −C5H11 group. Thus, pentanol is

partially soluble in water.



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