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:

$\pi =CRT$

$\pi$ = osmotic pressure = ?

C= concentration in Molarity  = 0.1 M

R= solution constant = 0.0821 Latm/Kmol

T= temperature = 300 K

$C_{glucose}=0.1\times 0.0821\times 300=2.5atm$

Isotonic solutions have same osmotic pressure at same temperatures.

Thus $\pi_{CaCl_2}=2.5atm$

$\pi_{CaCl_2}=i\times 0.04\times 0.0821\times 300$

$2.5=i\times 0.04\times 0.0821\times 300$

$i=2.5$

The dissociation reaction for $CaCl_2$ is:

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

$i=1+\alpha (n-1)$

i= vant hoff factor = 2.5

$\alpha$ = degree of ionization = ?

n = number of ions produced on ionization = 3

$2.5=1+\alpha (3-1)$

$\alpha=0.75$

$\alpha=0.75\times 100=75\%$

Thus the percentage of ionisation of $CaCl_2$ is 75%

Answer 2

Answer:

_____________ 75 % ___________