Question

(W g of a metal, on complete reaction with acid, liberates
V cm3 of H2, at S.T.P., the equivalent mass of the metal is
(a) W/V
(b) Wx11200/V
(c) V/W
(d)(W x 22.4)/V​

Answer 1

Answer:

Equivalent weight =valencymolecular weight

24=2Mw

Mw=48= Molecular weight of divalent metal X

Moles of X=4812=0.25

X+2HCl→XCl2+H2     

0.25     0.50          −         −

                           0.25       0.25

Moles of H2 produced =0.25 moles

Volume of 0.25 moles H2=0.25×22.4=5.6 litres