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Chemistry I

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Class 10 Class 12
Derive an expression relating the elevation of boiling point to the amount of solute present in the solution.

 Answer:

the vapour pressure of a liquid increases with increase of temperature. It boils at the temperature at which its vapour pressure is equal to the atmospheric pressure. For example, water boils at 373.15 K (100° C) because at this temperature the vapour pressure of water is 1.013 bar (1 atmosphere)

Let Tb0 be the boiling point of pure solvent and
Tb be the boiling point of solution. The increase in
the boiling point Tb = Tb - Tb0  is known as
elevation of boiling point.

for dilute solutions the elevation of boiling point (ΔTb) is directly proportional to the molal concentration of the solute in a solution. Thus
ΔTb ∝ m 
or ΔTb = Kb m
Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, Kb is called Boiling Point Elevation Constant or Molal Elevation Constant (EbullioscopicConstant). The unit of Kb is K kg mol-1. If w2 gram of solute of molar mass M2 is dissolved in w1 gram of solvent, then molality, m of the solution is given by the expression:

m = w2M2w11000  = 1000×w2M2×w1Substituting the value of molality in Tbm we get Tb = Kb×1000×w2M2×w1M2 =  Kb×1000×w2Tb×w1

Thus, in order to determine M2, molar mass of the solute, known mass of solute in a known mass of the solvent is taken and ΔTb is
determined experimentally for a known solvent whose Kb value is known.

1771 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 (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            
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 benzenemass of solution ×100= 2222+122×100= 22144×100 = 15.28%
Mass% of carbon tetrachloride = 100 - 15.28
                          = 84.72%
1703 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:

= 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

1475 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 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
Molarity, 
          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

                 
  

844 Views