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Class 10 Class 12
How is osmotic pressure of a solution determined ? If the membrane used was slightly leaky, how will it influence the measured value of osmotic pressure?
Osmotic pressure of a solution containing 7 g of a protein per 103 ml of solution is 25 mm Hg at 310 K. Calculate the molecular mass or the protein. (R = 0.0821 L atm K–1 mol).


Osmotic pressure: Osmotic pressure is the minimum pressure that should be applied to the more concentrated solution to prevent osmosis.
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus: Π = C R T 
Here Π is the osmotic pressure and R is the gas constant. 

(a) Measurement of Osmotic Pressure. Different methods are employed for the measurement of osmotic pressure in the laboratory but Berkley and Hartley's method gives the best results. The apparatus consists of a porous pot containing copper ferrocyanide deposited in its wall (acts as semi-permeable membrane) and fitted into a bronze cylinder to which is fitted a piston and a pressure gauge (to read the applied pressure).

The pot is fitted with a capillary indicator on left and water reservoir on right. Pot is filled with water while the cylinder is filled with a solution whose osmotic pressure is to be measured. Water tends to pass into the solution through the semipermeable membrane with the result that the water level in the indicator falls down. External pressure is now applied with piston so as to maintain a constant level in the indicator. This external pressure is osmotic pressure.
If the membrane used was a slightly, leaky, then the measured valued of osmotic pressure will not be definite.

Fig. Berkley and Hartley's apparatus.
we have given that
mB = 7g
R=0.0821 L atm K-1 mol

 MB = mBRTm       = 7×0.0821×3101001000×25760×54094 g mol-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 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 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 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