Question

0.04 M Cacl2, is isotonic with 0.1 M glucose
solution at 300 K temperature, then calculate the
percentage of ionisation of CaCl2,
(1) 55%
(2) 70%
(3) 60%
(4) 75%​

Answer 1

Answer:

(4) 75%

Explanation:

See attachment

Answer 2

Concentration in Molarity = C  = 0.1 M  (Given)

Solution constant = R = 0.0821 Kmol  (Given)

Temperature = T = 300 K  (Given)

Cglucose = 0.1 × 0.0821 × 300

= 2.5

πCacl2 = i × 0.04 × 0.0821 × 300  

2.5 = i × 0.04 × 0.0821 × 300  

i = 2.5

i = 1 + a ( n - 1)

where i is the vant off factor, a is the degree of ionization and n is the number of ions produced on ionization

2.5 = 1 + a ( 3-1)

a = 0.75

a = 0.75 100

= 75%

Thus, the percentage of ionization od Cacl2 is 75%.