Chapter Chosen


Book Chosen

Chemistry I

Subject Chosen


Book Store

Download books and chapters from 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
A solution is made by dissolving 30 g of a non-volatile solute in 90g of water. It has a vapour pressure of 2.8 kPa at 298 K. At 298 K, vapour pressure of pure water is 3.64 kPa. Calculate the molar mass of the solute.

The relative lowering of vapour pressure is given by the following expression. 
psolvent0 -psolutionpsolvent0 =n2n1+n2

Where psolvent0 is the vapour pressure of pure solvent, Psolution  is the vapour pressure of solution containing dissolved solute, n1 is the number of moles of solvent and n2 is the number of moles of solute.

For dilute solution n2<<n1, therefore the above expression reduces to 

psolvent0- psolutionpsolvent0 =n2n1= w2 xM1M2 x w1 .....(A)

Where w1 and w2 are the masses and M1 and M2 are the molar masses of solvent and solute respectively.

We are given that

W1 =30 g

W2 =30g

psolution =2.8 kpa

p0solvent =? And M2=?

subtituting these values in relation we get,psolvent0 -2.8psolvent0 =30 x18M2 x90psolvent0 -2.8psolvent0 =6M2  (1)similarly for second case we have the following value.w2 =90 gw1 =90+18 =108gpsolution =2.9 kpatherefore we getpsolvent0 -2.9psolvent0 =30 x18M2 x108 (2)dividing 1 and 2 we getpsolvent0 -2.8psolvent0-2.9 =65vapour pressure of water at 298 k is 3.4kpasubstituting the vaue of psolvent0 in 1 we get3.4 -2.83.4 =6M2= 34 gram Therefore mass of solute is 34 g .


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
=moles of  KImoles of KI + moles of water

=0.120.12+4.44= 0.0263            

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.


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

(a) Mol. mass of Co(NO3). 6H2O

               =58.9+(14+3×16)2+6(18)=58.9+(14+48)×2+108=58.9+124+108 = 290.9

Moles of Co(NO)3.6H2O
                                       =30290.9=0.103 mol.
Volume of solution = 4.3 L
          M=Moles of soluteVolume of solution in litre    = 1034.3 = 0.024 M

(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=0.5×301000mol =0.015mol
therefore molarity =0.015/0.5L 

thus molarity is 0.03M



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


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

= Moles of soluteMass of solvent in kg
Mol. mass of urea NH2CONH2
                 = 14 + 2 + 12 + 16 + 14 + 2
                 = 60 g mol-1

Molality (m) = Moles of soluteMass of solvent in kg

25 = Moles of solute2.5

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


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,


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 benzenemass of solution ×100= 2222+122×100= 22144×100 = 15.28%
Mass% of carbon tetrachloride = 100 - 15.28
                          = 84.72%