A g of a metal displaces V mL of H_2 at NTP. Equivalent weight E, of metal is:

MHT CET Chemistry question

g of a metal displaces mL of at NTP. Equivalent weight , of metal is:

Answer: (D)

By the law of equivalents E/ = A/(wt. of H₂), which is (a); since 22400 mL of H₂ weighs 2 g (1 mL weighs 0.0000897 g), wt. of

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Question ID VK-041304