consideram: n1 moli benzen , care contin 72n1 g C (6·12·n1)
n2 moli toluen " ' ' ' '' ' ' ' ' ' '' 84n2 g C (7·12·n2)
% C = (72n1+ 84n2) / (78n1+ 92n2) ·100 = 92
100·6(6n1 + 7n2) = 92(39n1 + 46n2)
150(6n1 + 7n2) = 23(39n1 + 46n2)
3n1 = 8n2
n1/8 = n2/3 = (n1 + n2)/11 = n/11 ⇒ n1 = 8n/11 n2 = 3n/11
%toluen = 300/11= 27,27%
% benzen = 800/11 = 72,72% (procente molare)
b) daca consideram m1 g benzen ( cu m1/78 ·6·12 = 12m1/13 g C) si
m2 g toluen (cu m2/92 ·7 ·12 = 21m2/23 g C) ⇒
⇒ (12m1/13 + 21m2/23) / (m1 + m2) = 92/100 = 23/25
300m1 / 13 + 525m2 / 23 = 23m1 + 23m2
m1/13 = 4m2/23 ⇒ m2 = 0,442m1 m1 + m2 = 1,442m1
% benzen = 100/1,442 = 69,35%
% toluen = 44,2/1,442 = 30,65% ( procente masice)
