$x$ gram of metal is deposited on passing $n$ electrons through its aqueous solution. The equivalent weight of the metal will be:

\frac{n}{x} \cdot N_{A}

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x . n . N_{A}

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\frac{x}{n} \cdot N_{A}

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x \cdot \frac{n}{N_{A}}

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Solution

The correct option is A $\frac{x}{n} \cdot N_{A}$
We know that,
Molecular weight $=$ Equivalent weight $\times$ $n$ factor
Mass of one molecule / atom $= \frac{M o l e c u l a r w e i g h t}{N_{A}}$
Here, mass of one atom $= x$
Hence, Equivalent weight $= \frac{x}{n} . N_{A}$

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x gram of metal is deposited on passing n electrons through its aqueous solution. The equivalent weight of the metal will be:

The correct option is A xn⋅NAWe know that,Molecular weight = Equivalent weight × n factorMass of one molecule / atom =MolecularweightNAHere, mass of one atom =xHence, Equivalent weight =xn.NA

$x$ gram of metal is deposited on passing $n$ electrons through its aqueous solution. The equivalent weight of the metal will be:

The correct option is A $\frac{x}{n} \cdot N_{A}$
We know that,
Molecular weight $=$ Equivalent weight $\times$ $n$ factor
Mass of one molecule / atom $= \frac{M o l e c u l a r w e i g h t}{N_{A}}$
Here, mass of one atom $= x$
Hence, Equivalent weight $= \frac{x}{n} . N_{A}$