A metal piece weighing  0.45 to 0.55 g when dissolved in dil. HCl produced 90 mL of H2 at NTP. If exact at. wt. of metal is 65, calculate the exact weight of the metal.

Let wt. of metal= 0.5 g; wt. of H2 at NTP. = 90 X 0.00009 g  Exact at. wt. of metal= 65; Eq. wt. of metal= E.  We  Wt. of metal  Wt. of H2 at NTP= Eq. wt. of metal  Eq. wt. of H2(=1.008)          …(i)0.590×0.00009=E1.008∴ E=0.5×1.00890×0.00009=62.2 Valency = At. wt.  Eq. wt. =6562.2≈1∴  Exact eq. wt. = Exact at. wt.  Valency =651=65. Let x= exact wt. of metal  From relation (1), we have:  x90×0.00009=651.008∴ x=65×90×0.000091.008 =0.5223g