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A non-volatile organic compound Z was used to prepare two solutions. Solution A contains 5.00 g of Z dissolved in 100 g of water and solution B has 2.31 g of Z dissolved in 100 g of benzene. Solution A has vapor pressure at the normal benzene boiling temperature. Calculate the molar mass of Z in solutions A and B and explain the difference. Answer: Sol. A: 24 g / mol; Sol. B: 248 g / mol. This difference can be attributed to the process of dimerization of the compound in benzene

A non-volatile organic compound Z was used to prepare two solutions. Solution A contains 5.00 g of Z dissolved in 100 g of water and solution B has 2.31 g of Z dissolved in 100 g of benzene. Solution A has vapor pressure at the normal benzene boiling temperature. Calculate the molar mass of Z in solutions A and B and explain the difference.
 
Answer: 
Sol. A: 24 g / mol; Sol. B: 248 g / mol. This difference can be attributed to the process of dimerization of the compound in benzene

Grade:6

1 Answers

Arun
25750 Points
6 years ago
 
Given data
  Mass of water = 100 g
  The mass of solute = 5 g
  Number of moles of water = mass / molar mass
                                          = 100g/18g/mol
                                       n2= 5.56 mol
  The vapor pressure of water at 100 deg C, P0 = 760mmHg
    The vapor pressure of solution at 100 deg C= 754.5 mmHg
   We know that
  Accroding to Raoults Law
    The vapor pressure of solution  =mole fraction of water *P0
               Molefraction of water = 754.5 mmHg / 760 mmHg
                                                  =0.9927
   Mole fraction of solute = 1-0.9927
                                      =0.00723
             Molefraction of solute =n1/(n1+n2) 
                    ∴ n1/(n1+n2) =0.00723
                        (n1+n2) /n1 = 1/0.00723
                          (n1+5.56mol) /n1  = 138.3
                                                 n1 = 0.04 mol
     Molar mass of solute Z = mass /moles
                                        = 5.0 g / 0.04 mol
                                        = 125 g/mol
 
CaseII
  Given data
  Mass of benzene = 100 g
  The mass of solute = 2.31 g
  Number of moles of benzene = mass / molar mass
                                            = 100g/78 g/mol
                                        n2 = 1.28 mol
  The vapor pressure of benzene at 100 deg C, P0 =760 mmHg
    The vapor pressure of solution at 100 deg C= 754.5 mmHg
   We know that
  Accroding to Raoults Law
    The vapor pressure of solution  =mole fraction of benzene *P0
               Molefraction of water = 754.5 mmHg / 760 mmHg
                                                  =0.9927
   Mole fraction of solute = 1-0.9927
                                      =0.00723
             Molefraction of solute =n1/(n1+n2) 
                    ∴ n1/(n1+n2) =0.00723
                        (n1+n2) /n1 = 1/0.00723
                          (n1+1.28mol) /n1  = 138.3
                                                 n1 = 0.00932 mol
     Molar mass of solute Z = mass /moles
                                        = 2.31 g / 0.00932 mol
                                        = 247.85 g/mol
      Molar mass of Z in case II isalmost double to the molar mass of Z in case I that means thesolute dimerizes in bezene.

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