(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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Some Basic Concepts of Chemistry

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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 .....

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$1 \ g$ of hydrogen combines with $80 \ g$ of bromine. $1 \ g$ of calcium (valency $= 2$) combines with $4 \ g$ of bromine. The equivalent weight of calcium is ...

Oxygen consists of $O^{16}, O^{17}$ and $O^{18}$ isotopes and carbon consists of isotopes of $C^{12}$ and $C^{13}$. Total number of $CO_2$ molecules possible are

The approximate atomic weight of an element is $26.89$. If its equivalent weight is $8.9$,the exact atomic weight of the element would be:

What is the equivalent mass of oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$?

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