Question

0.04M CaCl2 is isotonic with 0.1M glucose solution at 300K temperature then calculate the percentage of ionisation of CaCl2

Answer 1

Answer:

75%

Explanation:

See Attachment

Answer 2

Answer: 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 $\pi_{CaCl_2}=2.5atm$

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

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

$i=2.5$

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

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

i= vant hoff factor

$\alpha$ = degree of ionization

n = number of ions produced on ionization

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

$\alpha=0.75$

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

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