Q. Equivalent weight of $N H_{3}$ as a base is

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A

$17$

B

$17/3$

C

$1.7$

D

$17/2$

Solution:

$∴ N H_{3} + H^{+} \to N H_{4}^{+}$
As base, valence factor of $N H_{3} i s 1$
$∴ Eq . wt . = 17/1 = 17$