If 1.0 g of a metal combines with 8.89 g of | vedclass

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(B) The equivalent weight of a metal $(E_M)$ is calculated using the formula: $E_M = \frac{	ext{Weight of Metal}}{	ext{Weight of Bromine}} 	imes

Vedclass · Accepted Answer

$9$

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(B) The equivalent weight of a metal $(E_M)$ is calculated using the formula: $E_M = \frac{	ext{Weight of Metal}}{	ext{Weight of Bromine}} 	imes 	ext{Equivalent weight of Bromine}$ Given: Weight of metal $(W_M)$ = $1.0 \ g$ Weight of bromine $(W_{Br_2})$ = $8.89 \ g$ Equivalent weight of bromine $(E_{Br_2})$ = $80 \ g/eq$ Substituting the values: $E_M = \frac{1.0}{8.89} 	imes 80 \approx 8.998 \approx 9$ Therefore,the approximate equivalent weight of the metal is $9$.

If $1.0 \ g$ of a metal combines with $8.89 \ g$ of bromine,what is the approximate equivalent weight of the metal? (Given: Atomic weight of $Br = 80 \ g/mol$)

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