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NEET Class neet Answered

A solution M is prepared by mixing ethanol and H2O.The mole fraction of ethanol in mixture is 0.9     . Given: Kf H20-1.86 Kkg/mol ,Kf ethanol-2.0 Kkg/mol ,Kb H2O-0.52 Kkg/mol ,Kb ethanol-1.2 Kkg/mol, Standard freezing point of H20- 273K , Standard freezing point of ethanol- 155.7K , Standard boiling point of H2O- 373K , Standard boiling point of ethanol-351.5K ,vp of pure water -32.8mm Hg,vp of pure ethanol -40mm Hg, mol wt of H2O - 18g/mol, mol wt of ethanol- 46g/mol                                . Consider the solutions to be ideal dilute solutions and solutes to be non volatile and non-dissociative                                                                                                                                        .Ques 1 ) The fp of solution M is: (a) 268.7K ,(b) 268.5 K ,(c) 234.2 K , (d) 150.9 K                          .Ques 2 ) The vp of solution M is: (a) 39.3mmHg ,(b) 36.0mmHg ,(c) 39.5mmHg ,(d) 28.8mmHg    .Ques 3) Wter is added to the solution M such that the mole fraction of water in solution becomes 0.9. The bp of this solution is : (a)380.4K ,(b)376.2K, (c)375.5K ,(d)354.7K
Asked by Balbir | 30 Jul, 2019, 18:16: PM
answered-by-expert Expert Answer
Given:
 
Mole fraction of ethanol = 0.9
 
Mole fraction of water = 1-0.9 
 
                                = 0.1
 
Kf H2O = 1.86 K kg/mol
 
Kf ethanol = 2.0 K kg/mol
 
Kb H2O = 0.52 K kg/mol
 
Kb ethanol = 1.2 K kg/mol
 
Standard freezing point of H20 =273 K 
 
Standard freezing point of ethanol = 155.7 K 

Standard boiling point of H2O = 373 K

Standard boiling point of ethanol=351.5 K

Vapour pressure of pure water = 32.8 mm Hg

Vapour pressure of pure ethanol = 40 mm Hg

Molecular weight of H2O - 18 g/mol

Molecular weight of ethanol = 46 g/mol        
 
 
1) Freezing point of solution M,
 
We know,
 
increment straight T subscript straight f space equals space straight K subscript straight f space cross times space straight m

straight m equals space fraction numerator Mole space fraction space of space solute space cross times 1000 over denominator Mole space fraction space of space solvent space cross times Molar space mass space of space solvent space end fraction

space space space equals fraction numerator 0.1 over denominator 0.9 cross times 46 end fraction cross times 1000

space space space equals 2.415 space straight m

increment straight T subscript straight f space equals space 2 cross times 2.415

space space space space space space space equals space 4.83 space straight K
 
Freezing point of solution = 155.7 - 4.83
 
                                     =150.86 K
 
2) Vapour pressure of solution M is: 
 
P = P1X1 + P2X2
 
   = 32.8 × 0.1 + 40 × 0.9
 
   =3.28 +36
 
   = 39.28 mm Hg
 
   ≈ 39.3 mm Hg
 
3) Water is added to the solution M such that the mole fraction of water in solution becomes 0.9.
    The bp of this solution is,
 
We know,
 
Elevation in boiling point is
 
increment straight T subscript straight b space equals space straight K subscript straight b cross times straight m

straight m space equals space fraction numerator 0.1 cross times 1000 over denominator 0.9 cross times 18 end fraction

space space space space equals 6.17

increment straight T subscript straight b space equals 0.52 cross times 6.17

space space space space space space space space equals space 3.20 space straight K
 
Boiling point of solution = 373 + 3.20
 
                                  =376.20 K
Answered by Varsha | 31 Jul, 2019, 12:47: PM
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