0.426 g of a metallic oxide on heating in a current of hydrogen, liberated 0.12 g water. Calculate the equivalent weight of the metal.

wt. of metal oxide = 0.426 g;wt. of water = 0.12 g.g. mol. wt. of water, H2O=(2×1)+16=18g.(i) 18g H2O contain oxygen = 16g. 0,12gH2O contain oxygen =16×0.1218=0.107g∴    wt. of metal left behind =0.426-0.107 g = 0.319 g,But  wt. of metal  wt. of oxygen = Eq. wt. of metal  Eq. wt. of oxygen (=8)∴ Eq. wt. of metal = wt. of metal  wt. of oxygen ×8=0.319g×80.107g=23.8 Ans.