Question

The osmotic pressure of 0.1 M aqueous solution of $MgC{l}_2$ at 300 K is 4.92 atm. What will be the percentage ionization of the salt?

• A. $49\%$
• B. $29\%$
• C. $39\%$
• D. $69\%$

Answer 1

Answer: (A)

$49\%$

$\pi =iCRT$

$4.92=i\times 0.1\times 0.0821\times 300$

$\Rightarrow i=1.99$

$\alpha =\frac{i-1}{n-1}$

$\alpha =\frac{1.99-1}{3-1}=\frac{0.99}{2}=0.49$

Percentage ionization $=49\%$