The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid
(C6H5COOH) is dissolved in 25 g of benzene. If benzoic acid forms a dimer in
benzene, calculate the van’t Hoff factor and the percentage association of
benzoic acid. (Kf for benzene = 5.12 K kg mol-1)
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Kf=5.12,deltaTf=2.12,w1=25,w2=2.5,
M2=?
We knew the equation
M2= (kf×w2×1000)/(deltaTf ×w1)
M2= (5.12×2.5×1000)/(25×2.12)
By solving we get
M2=241.50
i (van’t Hoff factor)=original molecular mass ÷ calculated molecular mass
We knew that M1 (C6H5COOH)= 122 gm per mol.
Hence i= (M1/M2) = 122/241.50
M2=?
We knew the equation
M2= (kf×w2×1000)/(deltaTf ×w1)
M2= (5.12×2.5×1000)/(25×2.12)
By solving we get
M2=241.50
i (van’t Hoff factor)=original molecular mass ÷ calculated molecular mass
We knew that M1 (C6H5COOH)= 122 gm per mol.
Hence i= (M1/M2) = 122/241.50
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