--%>
Assignment Help - homework help

Explain solid in liquid solutions.

The French chemist Francois Marie Raoult (1886) carried out a series of experiments to study the vapour pressure of a number of binary solutions. On the basis of the results of the experiments, he proposed a generalization called Raoult's law which states that, 

The vapour pressure of a solution containing non-volatile solute is directly proportional to the mole fraction of the solvent.

In case of solution containing two components A (volatile solvent) and B (non-volatile solute) the vapour pressure of solution is given as

[Vapour pressure of solution] = [vapour pressure of solvent in solution (pA) ∝ [mole fraction of solvent (xA)]
                                                  
Or pA ∝ xA
                                                
Or, pA = kxA

Where k is proportionality constant.

For pure liquid, xA = 1 then k becomes equal to be vapour pressure of the pure solvent which is denoted by pA°.

Thus, pA = pA°xA                                            (i)

Or psolution × mole fraction of solvent.

For solutions obeying Raoult's law at all concentrations its vapour pressure would vary linearly from zero to the vapour pressure of pure solvent.

If mole fraction of solute is sB, then

xA + xB = 1 or xA = 1 - xB                             (ii)

From eqns. (i) and (ii),

pA = pA°(1 - xB) = pA° - pA°xB

Or, pA° - pA = pA°xB

2263_solids on liquids.png

   Related Questions in Chemistry

  • Q : Question relatede to calculate molarity

    Select the right answer of the question. What is molarity of a solution of HCl that contains 49% by weight of solute and whose specific gravity is 1.41 : (a) 15.25 (b) 16.75 (c) 18.92 (d) 20.08

    		•
  • Q : What is chemisorption or chemical

    When the forces of attraction existing between adsorbate particles and adsorbent almost of the same strength as chemical bonds, the adsorption is called chemical adsorption. This type of adsorption is also known as chemisorptions. Since forces of attraction existing b

    		•
  • Q : Hydroxide is highly insoluble in

     : 1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to

    		•
  • Q : Molecular basis of third law. The

    The molecular, or statistical, basis of the third law can be seen by investigating S = k in W.The molecular deductions of the preceding sections have led to the same conclusions as that stated in the third law of thermodynamics, namely, that a value can be

    		•
  • Q : Problem associated to vapour pressure

    Provide solution of this question. 60 gm of Urea (Mol. wt 60) was dissolved in 9.9 moles, of water. If the vapour pressure of pure water is P0 , the vapour pressure of solution is:(a) 0.10P0 (b) 1.10P0 (c) 0.90P0 (d) 0.99P0

    		•
  • Q : Problem on molecular weight of solid

    The vapor pressure of pure benzene at a certain temperature is 200 mm Hg. At the same temperature the vapor pressure of a solution containing 2g of non-volatile non-electrolyte solid in 78g of benzene is 195 mm Hg. What is the molecular weight of solid:

  • Q : Explain the process of coagulation of

    Presence of small concentrations of appropriate electrolyte is necessary to stabilize the colloidal solutions. However, if the electrolytes are present in higher concentration, then the ions of the electrolyte neutralize the charge on the colloidal particles may unite

    		•
  • Q : Sugar solution The solution of sugar in

    The solution of sugar in water comprises: (i) Free atoms (ii) Free ions (iii) Free molecules (iv) Free atom and molecules. Choose the right answer from the above.

    		•
  • Q : Ddd 4) The addition of S2- ion to

    4) The addition of S2- ion to Fe(OH)2(s). Explain why the addition of S2- ion to Cr(OH)3(s) does not result in the formation of Cr2S3(s).

    		•
  • Q : Explain Second Order Rate Equations.

    Integration of the second order rate equations also produces convenient expressions for dealing with concentration time results.A reaction is classified as second order if the rate of the reaction is proportional to the square of the concentration of one o

    		•