Question

0.1 M aqueous solution of mgcl2 at 300 k is 4.92 ATM. what will be the percentage ionization of the salt​

Answer 1

Given,

0.1M aqueous solution of MgCl₂ at 300K is 4.92 atm.

To find,

The percentage ionisation of the salt.

concentrated of aqueous solution, C = 0.1 M

Temperature, T = 300 K

osmotic pressure , π = 4.92 atm

using formula,

π = iCRT, where i is Van't Hoff factor.

⇒4.92 = i × 0.1 × 0.082 × 300

⇒i = 4.92/(0.1 × 0.082 × 300)

= 2

dissociation reaction of MgCl₂ is ...

MgCl₂ ⇔Mg²⁺ + 2Cl¯

1 - α α 2α

so, i = 1 - α + α + 2α = 1 + 2α

⇒2 = 1 + 2α

⇒α = 1/2 = 0.5

Therefore percentage ionisation is 50%

Answer 2

Explanation:

$\sf π = iCRT$

$\sf 4.92 = i×0.1×0.0821×300$

$\sf i = 1.99$

$\sf α = \dfrac{i-1}{n-1}=\dfrac{1.99-1}{3-1}$

$\sf = \dfrac{0.99}{2}=0.49$

$\sf Percentage \: of \: ionisation \: is \: 49 \%$