Question

Ag of a metal displaces $VmL$ of $H_{2}$ at $NTP$ . Equivalent mass $\left(\right.E\left.\right)$ of metal is :

• A. $E=\frac{A}{mass of H_{2} displaced}\times E_{H}$
• B. $E=\frac{A \times 1 . 008 \times 22400}{volume of H_{2} displaced \times 2}$
• C. $E=\frac{A \times 1 . 008}{volume of H_{2} displaced \times 0 . 0000897}$
• D. All of the above

Answer 1

Answer: (D)

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