1.

Vant hoff factor of BaCl_(2) of conc. 0.01 M is 1.98. Percentage dissociation of BaCl_(2) on this conc. Will be

Answer»

49
69
89
98

Solution :`{:(,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`.


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