The oxide of a metal was found to contain  52.94% metal. If V.D. of the chloride of the metal is 66.75, what will be the at. wt. of the metal?

Hint we know,  Wt. of metal  Wt. of oxygen = Eq. wt. of metal  Eq. wt. of oxygen  52.94g47.06g=E8E=9Mol. wt. or 2×x V.D. = Mol. wt. of MClx = M + 35.5 x2×66.75=Ex+35.5x=9x+35.5x=44.5x∴ x=2×66.7544.5=3∴  Exact at. wt. = Eq. wt. × Valency  = 9 x 3 = 27