Question

0.3 g of benzoic acid dissolved in 20 g of benzene then depression in freezing point is 0.317 k calculate degree of association of benzoic acid?​

Answer 1

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Answer 2

Given: Mass of benzoic acid = 20g, mass of benzene= 20 g, depression in freezing point= 0.317k

To find: Degree of association of benzoic acid

Solution: Let A be benzene and B be benzoic acid

we are given WA= 20g, WB= 0.3g, ΔTf= 0.317k

ΔTf= Kf× WB/ MB × 1000/WA

where Kf= Molal depression constant for benzene is 4.9 kg m−1

MB is the abnormal molecular mass

putting the values we will get MB that is

MB= 4.9×0.3× 1000/(0.317× 20) = 231.86 g/mol

we know that 2C6H5COOH⇌(C6H5COOH)2

Let the degree of dissociation be x, therefore (1-x) mole of benzoic acid will be left undissociated. And x/2 moles of C6H5COOH will be associated at equilibrium.

Therefore we can say that

Total number of moles that will participate in equilibrium= (1-x)+(x/2)= 1-(x/2) = i

i = Normal molecular mass/ abnormal molecular mass

1-(x/2)= 122/ 231.86 = 0.52617

x/2 = 0.47382

x= 0.947

Therefore, the degree of association of benzoic acid will be 0.947.