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:

Heya .....mate .....

Here's ur answer :

Here, isotonic solution means that their osmotic pressures are equal.

So, π = cRT

where, c = concentration of solute

R = Gas constant

T = Temperature

π = Osmotic pressure

.°. π(urea) = (3/60*0.1)*0.082*T

=> π(urea) = (0.082*T)/2

Now, KCl being on electrolyte , n = 2

----> i= ?

π(KCl) = i * (19/74.5*0.1)*0.082*T

Since, they are equal, we have

(19/7.45)*i*0.082*T = (0.082*T)/2

=> i = (74.5*10^-1)/(19*2)

=> i = 1.96*10^-1

Using degree of dissociation (α) = i-1/n-1

α = [(1.96 *10^-1)- 1]/(2 - 1)

α = 0.096

%α = 9.6%

❤✨Hope it helps✨❤