Question

Berapa fraksi mol 46 gram toluena (C7H8) dan 117 gram benzena (C6H6) dalam larutan?

Answer 1

Mol C₇H₈ = massa/Mr = 46/(7*12 + 8*1) = 46/92 = 0,5 mol
mol C₆H₆ = massa/Mr = 117/(6*12 + 6*1) = 117/78 = 1,5 mol

X total = X C₇H₈ + X C₆H₆
           = {(mol C₇H₈) / (mol C₇H₈ + mol C₆H₆)} + {(mol C₆H₆) / (mol C₆H₆ + mol C₇H₈)}
           = {(0,5) / (0,5 + 1,5)} + {(1,5) / (1,5 + 0,5)}
           = {(0,5)/2} + {(1,5)/2}
           = 2/2
           = 1