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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$9$

Vedclass · Answer

(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 $Br = 80 \ g/mol$)

Similar Questions

An element $X$ reacts with oxygen to form a compound $X_2O_3$. If the atomic mass of $X$ is $91.5$,the equivalent mass of $X$ is:

Heating $1.520 \ g$ of a metal hydroxide gives $0.995 \ g$ of the metal oxide. Calculate the equivalent weight of the metal.

Write about atomic mass.

$10 \, g$ of $CaCO_3$ contains

If $13.8 \, g$ of an element contains $4.6 \times 10^{22}$ atoms,what is the atomic mass of that element?

Vedclass Test Series

Exam Paper Generator

Online Exam Module