n1 moli C6H5-CH2Cl contine 35,5n1 g Cl M1 = 126,5 g/mol
n2 moli C6H5-CHCl2 " " " " " 71n2 g Cl M2 = 161 g/mol
35,5(n1 +2n2)·100/(126,5n1 + 161n2) = 37,68
3550n1+7100n2 = 4766,52n1 + 6066,48n2
1033,52n2 = 1216,52n1
n2 = 1,177n1
n1·100/(n1+n2) = 100n1/2,177n1 = 45,93% C6H5-CH2Cl