Question

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.

Answer 1

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%

Answer 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, $K_f$=4.9
use formula, ΔT = (1000 × $k_f$ × w)/(M × W)
where M is experimental molar mass of products
we get, M = 241.98g/mol

now, from theoritical ,
$nC_6H_5COOH\rightarrow(C_6H_5COO)_n$
after association, i = 1 - $\alpha+\frac{\alpha}{n}$
for dimerization, n = 2
i = $1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$
we know, from van Hoff's concepts ,
$\frac{M_{\textbf{theoritical}}}{M_{\textbf{experimental}}}=1-\alpha/2$
we know, $M_{\textbf{theoritical}}$ = 122g/mol

so, $\frac{122}{241.98}=1-\alpha/2$

or, $\alpha=0.992$

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