JEE MAIN - Chemistry (2016 - 10th April Morning Slot - No. 17)
An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is :
0.33
0.50
0.67
0.80
Explanation
Let us assume that degree of dissociation is $$\alpha$$.
$$\matrix{ {M{X_2}} & { \to {M^{2 + }} + } & {2X + } \cr {(1 - \alpha )} & \alpha & {2\alpha } \cr } $$
Thus, after dissociation total number of moles formed (n) = 3.
Now, we know degree of dissociation is
$$\alpha = {{i - 1} \over {n - 1}} = {{2 - 1} \over {3 - 1}} = 0.50$$
Comments (0)
