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

Question

The equivalent weight of P 4 in the following reaction is: P 4+ NaOH + H 2 O arrow PH 3+ NaH 2 PO 2 (where = M = molar mass) (A) ( M /3) (B) ( M /11) (C) ( M /6) (D) (M/2)

Tardigrade Admin · Accepted Answer

P 4+ NaOH + H 2 O arrow PH 3+ H 2 PO 2- Eq. wt. of P 4=( M / n text factor 1)+( M / n text factor 2) =(M/12)+(M/4) ⇒ (M+3 M/12)=(4 M/12)=(M/3)

Q. The equivalent weight of $P_{4}$ in the following reaction is:
$P_{4} + N a O H + H_{2} O \to P H_{3} + N a H_{2} P O_{2}$
(where $= M =$ molar mass)

$\frac{M}{3}$

$\frac{M}{11}$

$\frac{M}{6}$

$\frac{M}{2}$

$P_{4} + N a O H + H_{2} O \to P H_{3} + H_{2} P O_{2}^{-}$

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

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

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

Q. The equivalent weight of P4​ in the following reaction is: P4​+NaOH+H2​O→PH3​+NaH2​PO2​ (where =M= molar mass)

3M​

11M​

6M​

2M​