Question

$1.0g$ of metal combines with $8.89g$ of Bromine. Equivalent weight of metal is nearly:
(atomic weight of Br is $80$)

• A. $8$
• B. $9$
• C. $10$
• D. $7$

Answer 1

Answer: (B)

$9$

Let equivalent weight be $W$

$\therefore$ Equivalent of metal $=\frac{1}{W}$

Equivalent Weight $=\frac{Molecularweight}{n_{factor}}$

$n_{factor}\left(Br\right)=1$

Equivalent of $Br=\frac{8.89}{80/1}$

Equivalent of metal $=$ equivalent of $Br$

$\frac{1}{W}=\frac{8.89}{80}$

$W=8.99$