n1 moli CHCl3 m1 = 119,5n1; m1Cl = 106,5n1
n2 Moli CCl4 m2 = 154n2 ; m2Cl = 142n2
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m amestec = 119,5n1 + 154n2 mCl = 106,5n1 + 142n2
(106,5n1 + 142n2)·100/(119,5n1 + 154n2) = 91,477
10650n1 + 14200n2 = 10931,502n1+14087,458n2
281,502n1 = 112,542n2 2,5n1 = n2
%CHCl3 = n1·100/(n1 + n2) = 100n1/(3,5n1) = 28,57%
%CCl4 = 250/3,5 = 71,43% (procente molare)
%CHCl3 = m1·100/(m1 +m2) = 11950n1/[(119,5+385)n1]= 23,68%
%CCl4 = 76,32% procente masice