Q. 624.0( 1 Vote )

Why is the mass d

Answer :

1. We know that molecules of certain substances when dissolved in solvents either associate or dissociate. This behaviour is shown by ionic substances dissociating or associating in water.


2. If we take potassium chloride or KCl, dissolving it in water will cause it to dissociate into K+ and Cl- ions. This leads to the increase in the chemical particles in the solution compared to the particles of the pure solute added. Properties of solutions which depend on the number of solute particles in it are known as colligative properties. Osmotic pressure, lowering of vapour pressure, elevation of boiling point and depression of freezing point are colligative properties. So calculation of molar mass on the basis of colligative properties would be higher than experimental value in this case.


3. In the case of association of particles, ethanoic acid (acetic acid) molecules dissolved in benzene dimerize due to hydrogen bonding. The number of particles in the solution thus reduces. Here the molar mass calculated from colligative properties, will be lower than the experimental value.


4. A molar mass that is either lower or higher than the expected value deduced theoretically is called as abnormal molar mass.


5. The van’t Hoff factor, i, was introduced to account for the extent of association or dissociation. It can be defined as the ratio of the experimental colligative property to the calculated colligative property.


i = (Normal Molar Mass)/(Abnormal Molar Mass)


i = (Observed Colligative Property)/(Calculated Colligative Property)


i = (Total number of moles of particles after association/dissociation)/(Number of moles of particles before association/dissociation)


6. In the case of an association where the observed molar mass is more than normal, the value of i is less than unity. For dissociation where the observed molar mass is lesser than expected, it is greater than unity.


7. Inclusion of van’t Hoff factor in colligative properties modifies their equations as follows:


i. Relative lowering of the vapour pressure of solvent,


(p_1^0- p_1^o)/(p_1^0 )=i.n_2/n_1


ii. Elevation of Boiling point, ∆Tb = i.Kb.m


iii. Depression of Freezing point, ∆Tf = i.Kf.m


iv. Osmotic pressure of solution, π = i.n2.R.T/V


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