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Class 10 Class 12
Discuss the various types of plots between the partial vapour pressure and the mole fractions of two components of the completely miscible liquids in a solution.

Answer:

When the partial vapour pressures of different (two) miscible liquids are plotted against their compositions (mole fractions), following three types of vapour pressure-composition (p – x) curves are obtained.
(a) Type-I: When the vapour pressures of the mixture lie between the vapour pressure of pure components : In such cases the solution obeys the Raoult’s law (ideal solution) i.e., the partial vapour pressure of each component (pc) is obtained by the relation.

Fig. Solution obeying Raoult’s law.
Where xc is the mole fraction of that component and p0c the vapour pressure of that component in pure form. In such cases, p-x curve is always a straight line. The total vapour pressure of the solution is equal to the sum of the partial vaour pressures of all components.
Examples: Solution of benzene-toluene, chlorobenzene bromobenzene, hexane-heptane.
(b) Type-II : When the observed vapour pressure of the solution is greater than that of calculated vapour pressure from the Raoult’s law : In such cases partial vapour pressure of each component is found to be more than expected on the basis of the Raoult’s law. The total vapour pressure of the solution is also greater than the vapour pressure corresponding to the ideal solution. At a certain composition the total vapour pressure of the solution will be the highest (maximum) which is greater than the vapour pressure of either of the pure liquids (components) at this components the boiling point of the solution will be lowest. This type of deviation from Raoult’s law is known as positive deviation and the system exhibits a maximum value of vapour pressure at certain composition. At this composition both the liquids boil at same (constant) temperature (minimum boiling azeotropes). In figure point C, corresponds the composition of the two liquids which boils at lowest temperature. For example, alcohol-water mixture having the composition of 95.59% alcohol and 4.41 water boils at 78.130C. This composition is called azeotropic mixture.

Fig.p-x curve showing maximum in the total vapour pressure curve.
Example: Ethanol-water solution. Acetone-carbon disulphide solution. Chloroform-ethanol solution.
(c) Type-III: When the observed vapour pressure of the solution is less than that of calculated from Raoults law: In such cases the partial vapour pressure of any component is found to be less than the expected vapour pressure on the basis of Raoult’s law. Similarly, the total vapour pressure of the solution is also less than that of expected value according to Raoult’s law. At a certain composition the total vapour pressure of such solution will be lowest (minimum). At this composition the boiling point of the solution will be highest (maximum) and both the component will boil at same temperature without the change in the composition. Such composition corresponds to the maximum boiling azeotropic mixture. For example a mixture of 20.24% of HCl and 79.76% water forms an azeotropic mixture which boils at 1100C, without the change in composition. In figure, point C, corresponds the composition of azeotropic mixture. This type of deviation in Raoult’s law is known as negative deviation.
Example : Solution of water-HCl, chloroform-benzene, acetone-aniline.

Fig.  p-x curve showing minimum in the total vapour pressure curve.

153 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 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 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

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