## Book Store

Currently only available for.
CBSE Gujarat Board Haryana Board

## Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
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.

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

In order to account for all such abnormalities, introduced a factor
(i) known as van 't Hoff's factor, which represents the extent of association (or) dissociation of a solute.

Van't Hoff's factor is defined as the ratio of the observed colligative property to the calculated colligative property.

i = Observed colligative property / Calculated colligative property

Observed colligative property ∝ 1/Molar Mass

i = Mc/Mo

Van't Hoff's factor
(i) represents the extent of association (or) dissociation of a solute

i = Total number of moles of particles after association or dissociation / Number of moles of particles before association or disscussion

Experimentally determined molar mass is always lower than actual value for dissociation.

Molar Mass ∝ 1/Colligative Property

If the solute undergoes association in a solution, then the value of van 't Hoff's factor is less than one. If the solute undergoes dissociation then 'i' is greater than one.

Ex:

KCl → K+ + Cl-

1 molecule of KCl furnishes 2 ions in solution

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

i = 2/1 = 2

2CH3COOH ⇔(CH3COOH)2
Ethanoic acid     Dimer of Ethanoic acid

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

i = 1/2 =  0.5

237 Views

Calculate the mass of urea (NH2CONH2) required in making 2.5 kg 0.25 of molal aqueous solution.

Solution:

Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:

Mol. mass of urea ${\mathrm{NH}}_{2}{\mathrm{CONH}}_{2}$
= 14 + 2 + 12 + 16 + 14 + 2
=

Molality (m) =

or Moles of solute
= 0.25 x 0.25 =  0.625

Mass of urea
= Moles of solute x Molar mass

= 0.625 x 60 = 37.5 g

1475 Views

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.

(a) 20% (mass/mass) means that 20 g of KI is present in 80 g of water.

Therefore, Moles of KI in solution

moles of KI = 20/166 =0.12mol
moles of water =80/18 =4.44mol
therefore, mole fraction of KI

=

1010 Views

Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Mass % of benzene

Mass% of carbon tetrachloride = 100 - 15.28
= 84.72%
1703 Views

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Let the total mass of the solution be 100g and mass of benzene be 30 g
therefore mass of tetrachloride= (100-30)g = 70g
Molar mass of benzene,

897 Views

Calculate the molarity of each of the following solution (a) 30 g of Co(NO3)2.6H2O in 4.3 L solution (b) 30 mL of 0.5 MH2SO4 diluted to 500 mL.

solution;

Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

(a) Mol. mass of

Moles of $\mathrm{Co}\left(\mathrm{NO}{\right)}_{3}.6{\mathrm{H}}_{2}\mathrm{O}$

Volume of solution = 4.3 L
Molarity,

(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=$\frac{0.5×30}{1000}$mol =0.015mol
therefore molarity =0.015/0.5L

thus molarity is 0.03M

844 Views