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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`
one year ago

```							Dear student  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.  RegardsArun (askIITians forum expert)
```
one year ago
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