Aman Bansal
Last Activity: 12 Years ago
Dear Dev
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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Thanks
Aman Bansal
Askiitian Expert
