A) 35.54
B) 36.54
C) 17.77
D) 18.27
Correct Answer: B
Solution :
\[\frac{Ew\,of\,acid}{Ew\,of\,salt\,of\,Mg}\]=\[\frac{weight\,of\,acid}{Weight\,of\,slt}\] Mg+2HCl (monobasic acid)\[\to \]\[MgC{{l}_{2}}\]+\[{{H}_{2}}\] Let Ew of acid=Ew of H + Ew of acid radical \[\therefore \]Ew of salt of Mg = Ew of Mg + Ew of acid radical \[\therefore \]\[\frac{Ew\,of\,acid}{Ew\,of\,Mg\,salt}\]=\[\frac{weight\,of\,acid}{Weight\,of\,Mg\,salt}\] \[\Rightarrow \]\[\frac{Ew\,of\,H+Ew\,of\,acid\,radical\,(E)}{Ew\,of\,Mg\,+Ew\,of\,acid\,radical\,(E)}\]=\[\frac{1.0}{1.301}\] \[\Rightarrow \] \[\frac{1+E}{12+E}=\frac{1.0}{1.301}\] \[\therefore \] E=35 , 54 \[\therefore \] EW of acid = Ew of H + Ew of acid radical =1+35.54=36.54
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