Question

Un amestec de n-C4H9Cl si i-C3H7Cl contine 40%,in procente de masa,clor.Determinati procentele masice,si apoi cele molare ale celor doua substante aflate in amestec.

Answer 1

n1 moli  C4H9Cl     M1 = 92,5 g/mol   n1 moli Cl       m1Cl = 35,5n1 g
n2 moli  C3H7Cl     M2 = 78,5 g/mol    n2 moli Cl       m2Cl = 35,5n2 g
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amestec : m = 92,5n1 + 78,5n2    mCl = 35,5(n1 + n2)
35,5·100(n1+ n2) /(92,5n1 + 78,5n2) = 40
88,75n1 + 88,75n2 = 92,5n1 + 78,5n2
10,25n2 = 3,75n1   41n2 = 15n1   n2 = 15/41 ·n1
%C4H9Cl = 92,5n1·100/(92,5n1+ 78,5n2) = 9250·41/(3792,5+1177,5) =
= 379250/4970= 76,3%     %C3H7Cl = 23,7%  (procente de masa)
%C4H9Cl = n1·100/(n1+n2) = 100·41·/56 = 73,21%; %C3H7Cl = 26,79%  (procente molare)