Chemistry, asked by mansimarch4372, 1 year ago

Give reason when 2g of benzoic acid is dissolved in 25g of benzene, the experimentally determined molar mass is always greater than the true value

Answers

Answered by shivakvsuju148

First we take

W=2kg k=4.9 kg/mol

w1=25g ΔTf=1.62k

substituting this in the equation.

Δt=(k*w2*1000)/m2*w1

m2=241.98 g/mol

eperimental molar mass of benzoic acid=241.98 g/mol

now we need consider.

2 C6H5COOH---->(C6H5C00H)2

X represents the degree of association of solute the (1-x) mol of bezoic left and x/2.

1-x+x/2 = 1-x/2 .

i= normal molecular mass/abnormal molecular mass

x/2=1-122/241.98

x=2*0.496 =0.992

so the degree of association is 99.2%

when 2g of benzoic acid is dissolved in 25g of benzene , the experimentally determined molar mass is always greater than the true value this is because benzoic acid will dimerize.

Let's solve mathematically to understand better.

weight of benzoic acid, w = 2g

weight of benzene , W = 25g

change in freezing point , ∆T = 1.62

coefficient of freezing point, =4.9

use formula, ΔT = (1000 × × w)/(M × W)

where M is experimental molar mass of products

we get, M = 241.98g/mol

now, from theoritical ,

after association, i = 1 -

for dimerization, n = 2

i =

we know, from van Hoff's concepts ,

we know, = 122g/mol

so,

or,

here it is clear that, association constant is less than 1 that's why experimental molar mass is greater than the true value.