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Asked by yash01298 | 26 Jul, 2019, 07:49: PM
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Given:
 
Wsolute =19.5 gm
 
Wsolvent = 500 gm
 
ΔTf = 1 ºC
 
Kf = 1.86 K kg/mol
 
Molecular weight can be calculated by using formula;
 
straight M subscript 2 space equals space fraction numerator straight K subscript straight f cross times straight W subscript solute cross times 1000 over denominator increment straight T subscript straight f cross times straight W subscript solvent end fraction space space space space space equals space fraction numerator 1.86 cross times 19.5 cross times 1000 over denominator 1 cross times 500 end fraction straight M subscript 2 space equals space 72.54 space straight g divided by mol
 
Calculated molecular weight of CH2FCOOH is,
 
12+(1×2) + 19 +12+(16×2)+1 = 78 g/mol
 
The van't Hoff factor is ,
 
i = equals space straight M subscript cal over straight M subscript obs equals space fraction numerator 78 over denominator 72.54 end fraction equals 1.075
 
To calculate dissociation constant,
 
The equation is,
 
space space space space space space space space space space space space CH subscript 2 FCOOH space space space rightwards harpoon over leftwards harpoon with space space space space space space space space on top space CH subscript 2 FCOO to the power of minus space space plus space space space straight H to the power of plus space space space Initial space space space space space space space space space space straight C thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space space space space space space space space space space space space space space space space 0 space space space space space space space space space space space space space space space space space 0 At space eqm space space space straight C left parenthesis 1 minus straight alpha right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space Cα space space space space space space space space space space space space space space Cα
 
Total moles = C -Cα +Cα+Cα
 
                  =C(1+α)
 
We know,
 
van't Hoff factor, i = space equals fraction numerator straight C open parentheses 1 plus straight alpha close parentheses over denominator straight C end fraction space space equals space 1 plus straight alpha straight alpha equals space straight i minus 1
 
We have, i = 1.075
 
So, α = 1.075 - 1
 
         = 0.075
 
Degree of dissociation = 0.075
 
Concentration, C is calculated as,
 
equals fraction numerator No. space of space moles over denominator Volume space in space straight L end fraction equals fraction numerator begin display style bevelled fraction numerator 19.5 over denominator 78 end fraction end style over denominator 0.5 end fraction equals 0.5 space straight M
 
Dissociation constant, Ka is
 
equals fraction numerator open square brackets CH subscript 2 FCOO to the power of minus close square brackets open square brackets straight H to the power of plus close square brackets over denominator open square brackets CH subscript 2 FCOOH close square brackets end fraction equals fraction numerator Cα squared over denominator 1 minus straight alpha end fraction equals fraction numerator 0.5 cross times 0.075 over denominator 1 minus 0.075 end fraction equals 3.06 cross times 10 to the power of negative 3 end exponent
 
 
van't Hoff factor is 1.075
Dissociation constant is 3.06×10-3 
 
Answered by Varsha | 29 Jul, 2019, 11:35: AM

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