Ka, the acid dissociation constant, for an acid is 9 × 10^{−4} at room temperature. at this temperature, what is the approximate percent dissociation of the acid in a 1.0 m solution?
(a) 0.03%
(b) 0.09%
(c) 3%
(d) 5%
(e) 9%
can you explain with some details?

Dissociation=k×no of moles
percentage of dissociation=9.0×10^-4×1×100
knowing that x%=x/100,we then say;
x/100=9.0×10^-4×1×100
therefore, x=100×100×9×10^-4×1
x=9
x percentage of dissociation=9%

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