Q. Ag of a metal displaces $Vm L$ of $H_{2}$ at $NTP$ . Equivalent mass $(E)$ of metal is :

208 153 NTA AbhyasNTA Abhyas 2020 Report Error

A

$E = \frac{A}{ma sso f H _{2} d i s pl a ce d} \times E_{H}$

B

$E = \frac{A \times 1.008 \times 22400}{v o l u m eo f H _{2} d i s pl a ce d \times 2}$

C

$E = \frac{A \times 1.008}{v o l u m eo f H _{2} d i s pl a ce d \times 0.0000897}$

D

All of the above

Solution:

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