The same amount of electricity is passed through acidified water and an aqueous solution of a metal chloride . As a result 7.4L `H_(2)` , gas at STP in the first case and 21 g metal in the second case are produced . Calculate the equivalent weight of the metal ? If the specific heat of the metal is 0.094 , then find the atomic weight of the metal ?

Explanation

`H^(+)+erarr(1)/(2)H_(2)`, Hence, `"11.2 L "H_(2)" (NTP)"-=96500C`
Thus, `63758.92C-="21 g metal"`
`therefore" 96500 C"-="31.78 metal."`
`therefore" Equivalent mass of the metal "=31.78`
But, specific heat `xx` atomic mass (approx)=6.4
`therefore" Atomic mass (approx)"=(6.4)/(0.094)=68.08,` valency `=(68.08)/(31.78)~~2`
`therefore" Accurate atomic mass "=2xx31.78=63.56`