1.

Vant hoff factor of `BaCl_(2)` of conc. 0.01 M is 1.98. Percentage dissociation of `BaCl_(2)` on this conc. Will beA. 49B. 69C. 89D. 98

Answer» Correct Answer - A
`{:(,BaCl_(2),hArr,Ba^(2+),+,2Cl^(-)),("Initially",1,,0,,0),("After dissociation",1-alpha,,alpha,,2alpha):}`
Total `= 1 - alpha + alpha + 2alpha = 1 + 2 alpha`
`alpha = (i-1)/(m-1) = (1.98 - 1)/(3 - 1) = (0.98)/(2) = 0.49`
m = number of particles in solution
For a mole `alpha = 0.49`
For 0.01 mole `alpha = (0.49)/(0.01) = 49`.


Discussion

No Comment Found

Related InterviewSolutions