0.32 g of metal gave on treatment with an acid 112 mL of hydrogen at NTP. Calculate the equivalent weight of the metal.

0.32 g of metal gave on treatment with an acid 112 mL of hydrogen at N

1.	0.32 g of metal gave on treatment with an acid 112 mL of hydrogen at NTP. Calculate the equivalent weight of the metal.
Answer» Solution :Mass of the metal taken (`w_(1)) = 0.65` g Mass of 224 ML of hydrogen LIBERATED `w_(2) = 2/(22400) XX 224 = 0.02 g` HENCE, Equivalent weight of the metal `=w_(1)/w_(2) =0.65/0.02 = 32.5`