The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formation of product A. The reaction 1 mol of A with excess of AgNO$$_3$$ in aqueous medium gives ________ mole(s) of Ag. (Round off to the Nearest Integer).

When white phosphorus ($$P_4$$) is boiled with concentrated NaOH solution in an inert atmosphere, it undergoes disproportionation: $$P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2$$. Two products are formed — phosphine gas ($$PH_3$$) and sodium hypophosphite ($$NaH_2PO_2$$). The product A referred to in the question is the hypophosphite salt $$NaH_2PO_2$$, which remains in the aqueous solution.

In $$NaH_2PO_2$$, phosphorus is in the +1 oxidation state. Hypophosphite is a powerful reducing agent. When 1 mol of $$NaH_2PO_2$$ reacts with excess $$AgNO_3$$ in aqueous medium, the phosphorus gets oxidized from the +1 state to the +5 state (as phosphate), losing 4 electrons per P atom. Each $$Ag^+$$ gains 1 electron to form metallic Ag. Therefore, the balanced reaction is: $$NaH_2PO_2 + 4AgNO_3 + 2H_2O \to NaH_2PO_4 + 4Ag + 4HNO_3$$.

Hence, 1 mol of $$NaH_2PO_2$$ gives 4 moles of Ag. The answer is 4.