19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
Molecular mass of CH2FCOOH
14 + 19 + 12 + 16 + 16 + 1 = 78 g/mol
Now, Moles of CH2FCOOH = 19.5 / 78
= 0.25
Taking the volume of the solution as 500 mL, we have the concentration:
C = (0.25 / 500) X 1000
Therefore Molality = 0.50m
So now putting the value in the formula :
ΔTf = Kf x m
=1.86 x 0.50 = 0.93K
Van’t hoff factor = observed freezing point depression / calculated freezing point depression
= 1 / 0.93 = 1.0753
Let α be the degree of dissociation of CH2FCOOH
Now total number of moles = m(1-a) + ma +ma = m(1+a)
Or
i = α(1+α) / α = 1 +α = 1.0753
Therefore α = 1.0753- 1
= 0.0753
Now the Value of Ka is given as:
Ka = [CH2FCOO-][H+] / CH2FCOOH
= (Cα x Cα) / (C (1-α))
= Cα2 / (1-α)
Ka = 0.5 X (0.0753)2 / (1-0.0753)
= 0.5 X 0.00567 / 0.09247
= 0.00307 (approx.)
= 3 X 10-3
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(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone
(vii) Ethanal and Propanal
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
How the following conversions can be carried out?
(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
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Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanonickelate(II)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaamminecobalt(III) sulphate
(vii) Potassium tri(oxalato)chromate(III)
(viii) Hexaammineplatinum(IV)
(ix) Tetrabromidocuprate(II)
(x) Pentaamminenitrito-N-cobalt(III)
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Convert
(i) 3-Methylaniline into 3-nitrotoluene.
(ii) Aniline into 1,3,5-tribromobenzene.
Why do we take, K(f) to be 1.86?
Value of c will be equals to molality
Great job
Great job
What a explanation
How the value of C is calculated for Ka...???
C can be calculated by c=n/v
How the value of 'C' is calculated for Ka value.
Great hard work.