(C) For a divalent metal $M$,the valency factor $(V.F.)$ is $2$.Equivalent weight $(E_w)$ = $\frac{\text{Atomic weight } (A_w)}{V.F.}$$32.7 = \frac{A_

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(C) For a divalent metal $M$,the valency factor $(V.F.)$ is $2$.Equivalent weight $(E_w)$ = $\frac{\text{Atomic weight } (A_w)}{V.F.}$$32.7 = \frac{A_w}{2} \implies A_w = 32.7 \times 2 = 65.4 \ g/mol$.The formula of the chloride of a divalent metal is $MCl_2$.Molecular weight $(M_w)$ = $A_w(M) + 2 \times A_w(Cl)$$M_w = 65.4 + 2 \times 35.5 = 65.4 + 71.0 = 136.4 \ g/mol$.

Class 11 Chemistry Some Basic Concepts of Chemistry Atomic, Molecular and Equivalent masses

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The equivalent weight of a divalent metal is $32.7$. The molecular weight of its chloride is .....

A
$68.2$
B
$103.7$
C
$136.4$
D
$166.3$

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Class 11

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Some Basic Concepts of Chemistry

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Atomic, Molecular and Equivalent masses

Atomic weight of an element is $x$. The actual mass of one atom of that element is

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If $1.2 \ g$ of a metal displaces $1.12 \ L$ of hydrogen at $NTP$,what is the equivalent weight of the metal?

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The relative abundance of two isotopes of atomic weight $85$ and $87$ is $75\%$ and $25\%$ respectively. The average atomic weight of the element is:

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The oxide of a metal contains $40 \%$ of oxygen. The valency of the metal is $2$. What is the atomic weight of the metal?

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If a metal oxide contains $32\%$ oxygen,what is the equivalent mass of the metal?

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Similar Questions

Atomic weight of an element is $x$. The actual mass of one atom of that element is

If $1.2 \ g$ of a metal displaces $1.12 \ L$ of hydrogen at $NTP$,what is the equivalent weight of the metal?

The relative abundance of two isotopes of atomic weight $85$ and $87$ is $75\%$ and $25\%$ respectively. The average atomic weight of the element is:

The oxide of a metal contains $40 \%$ of oxygen. The valency of the metal is $2$. What is the atomic weight of the metal?

If a metal oxide contains $32\%$ oxygen,what is the equivalent mass of the metal?

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