Question

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

• A. $49\%$
• B. $59\%$
  
• C. $79\%$
  
• D. $69\%$

Answer 1

The correct option is A $49\%$

$\pi =icRt$

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

$\pi =icRt$

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

$i=1.99$

$\alpha =\frac{i-1}{n-2}=\frac{1.99-1}{3-1}=\frac{99}{2}=49$(approxiamately)

Therefore, $\%$ of ionization = $49\%$