Dissolving 1.24 g of white phosphorous in boiling NaOH solution in an inert atmosphere gives a gas Q. The amount of CuSO$$_4$$ (in g) required to completely consume the gas Q is ______.

[Given : Atomic mass of H = 1, O = 16, Na = 23, P = 31, S = 32, Cu = 63]

White phosphorous is $$P_4$$.  In a boiling, strongly alkaline medium it undergoes disproportionation to give phosphine (the gaseous product Q) and sodium hypophosphite:

$$P_4 + 3\,NaOH + 3\,H_2O \;\longrightarrow\; 3\,NaH_2PO_2 + PH_3 \uparrow$$

Step 1 : Moles of $$P_4$$ taken
Molar mass of $$P_4 = 4 \times 31 = 124\,$$g mol$$^{-1}$$.
$$n(P_4)=\frac{1.24\;\text{g}}{124\;\text{g mol}^{-1}} = 0.01\;\text{mol}$$

Step 2 : Moles of phosphine ($$PH_3$$) formed
From the above equation, 1 mol $$P_4$$ gives 1 mol $$PH_3$$. Therefore
$$n(PH_3)=0.01\;\text{mol}$$

Step 3 : Reaction of $$PH_3$$ with $$CuSO_4$$
In an aqueous $$Cu^{2+}$$ solution, phosphine reduces the cupric ion, giving a black precipitate of copper phosphide $$Cu_3P_2$$. The stoichiometric equation is

$$2\,PH_3 + 3\,CuSO_4 + 3\,H_2O \;\longrightarrow\; Cu_3P_2\!\downarrow + 3\,H_2SO_4 + 3\,H_2$$

Thus, 2 mol $$PH_3$$ consume 3 mol $$CuSO_4$$, i.e. $$\dfrac{3}{2}=1.5$$ mol $$CuSO_4$$ per mol $$PH_3$$.

Step 4 : Moles of $$CuSO_4$$ required
$$n(CuSO_4)=1.5 \times n(PH_3)=1.5 \times 0.01=0.015\;\text{mol}$$

Step 5 : Mass of $$CuSO_4$$
Molar mass of $$CuSO_4=63+32+4\times16=159\,$$g mol$$^{-1}$$.
$$m(CuSO_4)=0.015 \times 159 = 2.385\;\text{g}$$

On rounding to two decimal places, the mass lies between 2.38 g and 2.39 g.

Hence, the amount of $$CuSO_4$$ required is 2.38 - 2.39 g.