The vapor pressure of pure water at 100°C is normally 760.0 mmHg, but decreased to 668.2 mmHg upon addition..

The vapor pressure of pure water at 100°C is normally 760.0 mmHg, but decreased to 668.2 mmHg upon addition of an unknown amount of copper (II) nitrate, Cu(NO3)2, to 216.5 g of water at this temperature. How many grams of copper (II) nitrate were added? Assume complete dissociation of solute.
The answer is 94.61

but i got
668.2/760 = Xa
216.5/18= 12.03 mol water
.8792= 12.03/ (12.03 + x)
10.58 + x= 12.03
x= 1.45 / 2 (dissociation)
.725 x 187.5563 = 135.97

how do i get 94.61?

216.5 / 18.015 = 12.018 mol water

668.2 / 760 = 0.8792 is the mole fraction that is water:

0.8792 = 12.018 / (12.018 + x)
find moles of ions released from Cu(NO3)2 , "x"

0.8792 (12.018 + x) = 12.018
10.566 + 0.8792x = 12.018
0.8792x = 12.018 - 10.566
0.8792x = 1.4516
x = 1.651 moles of ions released from Cu(NO3)2

since Cu(NO3)2 releases 3X's as many moles of ions:
Cu(NO3)2 (s) --> Cu+2 (aq) & 2 (NO3)-1 (aq)
so
1.651 moles of ions was released from ,1/3rd as many moles = 0.5504 moles of Cu(NO3)2

find grams, using molar mass:
(0.5504 moles of Cu(NO3)2 ) (187.5563 g/mol) = 103.2 grams of Cu(NO3)2

i don't see how you can get the 94.61