Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol–1. What is the percentage association of acid if it forms dimer in solution?

2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol–1. What is the percentage association of acid if it forms dimer in solution?

Grade:11

2 Answers

Meera Nair
11 Points
4 years ago
W= 2g; Kf = 4.9K kg mol-1; W1 = 25g; deltaTf = 1.62K
Substituting this in the equation  deltaTf = (Kf  x W2  x 1000) / (M2 x W1)
Therefore, M2 = (4.9 x 2 x 1000) / (25 x 1.62)    which is equal to 241.98 g mol-1
Thus experimental molar mass of benzoic acid in benzene is = 241.98 g mol-1
 Now consider the following equilibrium for the acid :  2C6H5COOH -------> (C6H5COOH)2
If x represents the degree of association of the solute then we would have (1 – x) mol of benzoic acid left in unassociated form and correspondingly x/2 as associated moles of benzoic acid at equilibrium is : 1 – x + x/2 = 1 – x/2
Thus, total number of moles of particles at equilibrium equals van’t Hoff factor i,
But i = Normal molar mass / Abnormal molar mass
        = 122 g mol-1 / 241.98 g mol-1
Or x/2 = 1 – 122/ 241.98 
          = 1 – 0.504
          = 0.496
Or x    = 2 x 0.496
          = 0.992
Therefore, degree of association of benzoic acid in benzene is 99.2%.
Yash Chourasiya
askIITians Faculty 256 Points
9 months ago
Dear Student

Please see the solution in the attachment.
643-2016_Untitled.png
I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free