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sorry for the last incomplete question the original one is: A non volatile organic compound is used to make two solutions A,B .Solution A contains 5 gm of Z dissolved in 100 gm of water and solution B contains 2.31 gm of Z in 100 gm of benzene .Solution A has a vapour pressure of 754.5 mm Hg at the normal boiling point of water and solution B has the same vapour pressure at the normal boiling point of benzene.calculate the molar mass of Z in the solutions A and B and account for the difference
sorry for the last incomplete question the original one is:
A non volatile organic compound is used to make two solutions A,B .Solution A contains 5 gm of Z dissolved in 100 gm of water and solution B contains 2.31 gm of Z in 100 gm of benzene .Solution A has a vapour pressure of 754.5 mm Hg at the normal boiling point of water and solution  B has the same vapour pressure at the normal boiling point of benzene.calculate the molar mass of Z in the solutions A and B and account for the difference

```
8 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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```
8 years ago
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