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Class 10 Class 12
A sample of drinking water was found to be severely contaminated with chloroform (CHCl3), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.


(i) 15 ppm of CHCl3 in water means that there is 15 g of CHCl3 in 106 g of water (1 million = 106)
Percentage of CHCl3
= 15106×100 = 0.0015 % = 1.5 × 10-4%

Here we have already find the mass % is 15 x 10-4 

When even mass % is given take total mass of solution = 100 gram
And mass of solute (CHCl3)  will  =  15 x10-4 g
Mass of solvent = mass of solution – mass of solute
(Here mass of solute is very small as compare to total mass of solution so neglect it and we get)
Mass of solvent = mass of solution
Mass of solvent = 100 g

 Number of moles of solute = mass of solute / molar mass

number of moles of CHCl3 =15×10-4118.5 = 1.266×10-5

Molality, m of CHCl3 in drinking water sample
            = moles of CHCl3mass of water in kg

Now plug the value of number of moles and mass of solvent in above eqution
Here mass of solvent = 100 g
Convert it in Kg we divide by 100 we get
Mass of solvent = 0.1Kg 
molaity = 1.266 x 10-5 / 0.1 = 1.266 x 10-4 mol Kg-1


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 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%

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