Question

The equivalent weight of $P _{4}$ in the following reaction is:
$P _{4}+ NaOH + H _{2} O \rightarrow PH _{3}+ NaH _{2} PO _{2}$
(where $= M =$ molar mass)

• A. $\frac{ M }{3}$
• B. $\frac{ M }{11}$
• C. $\frac{ M }{6}$
• D. $\frac{M}{2}$

Answer 1

Answer: (A)

$P _{4}+ NaOH + H _{2} O \rightarrow PH _{3}+ H _{2} PO _{2}^{-}$

Eq. wt. of $P _{4}=\frac{ M }{ n \text { factor }_{1}}+\frac{ M }{ n \text { factor }_{2}}$

$=\frac{M}{12}+\frac{M}{4} $

$\Rightarrow \frac{M+3 M}{12}=\frac{4 M}{12}=\frac{M}{3}$