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%

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.

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.

Answer 2

Answer:

Explanation:2g of benzoic acid dissolved in 25 g of benzene मे घोलने पर हिमांक मे 1.62K का अवनमन होता benzione के लिए मोलल अवनमन स्थिरांक 4.9Kkg mol–1 है यदि यह विलयन dimer बनाता है तो अम्ल का सगुणन कितना % होगा