Question

A solution containing 19 g per 100 mL of KCI (M 74-5 g mol-1) is isotonic with a solution containing 3 g per 100 mL of urea (M 60 g mol-1). Calculate the degree of dissociation of KCI sol Assume that both the solutions temperature. ​

Answer 1

Hey mate,

● Answer -
Degree of dissociation is 9.8 %.

● Solution -
Osmotic pressure of urea is calculated by -
P = nRT/V
P = 3/60 × 8.314 × 300 / 10^-4
P = 1.25×10^6 Pa

As KCl soln is isotonic to urea solution, both must be having same osmotic pressure.
P = 2αnRT/V
1.25×10^6 = 2 × α × 19/74.5 × 8.314 × 300 / 10^-4
α = 0.098
α = 9.8 %

Therefore, degree of dissociation is 9.8 %.

Hope this helped you...

Answer 2

Answer:

$9.8\%$

Explanation:

Osmotic pressure of urea is calculated by

P= nRT/V

P = 3/60 x 8.314 x 300 / 10^-4

P = 1.25 × 10^6 Pa

As KCI soln is isotonic to urea solution, both must be having same osmotic

pressure. P = 2anRT/V

1.25*10^6= 2 × ax 19/74.5 8.314 300 /10^-4

a = 0.098

a = 9.8 %

Therefore, degree of dissociation is 9.8%.