Question

Dissolving $1.24\text{g}$ of white phosphorous in boiling $\text{NaOH}$ solution in an inert atmosphere gives a gas Q. The amount of ${\text{CuSO}}_4$ (in $\text{g}$ ) required to completely consume the gas $\text{Q}$ is .
[Given : Atomic mass of $\text{H}=1,\text{O}=16,\text{Na}=23,\text{P}=31,$ $\text{S}=32,\text{Cu}-63$ ]

Answer 1

$\begin{array}{c}{P}_4+\text{3NaOH +}3H_{2}O⟶\stackrel{{}^{\prime }{Q}^{\prime }}{{\text{PH}}_3}+3{\text{NaH}}_2{\text{PO}}_2\\ Molesofphosphorous=1.24/124=0.01mol\\ \text{Since 1 mol of P forms 1 mol of}P{H}_3\text{, 0.01 mol of P will form 0.01 mol of}P{H}_3\\ 2{\text{PH}}_3+3CuS{O}_4⟶C{u}_3{P}_2+3H_{2}S{O}_4\end{array}$

No. of moles of $CuS{O}_4$ required to consume 2 mol of phosphine (Q) = 3
So, no. of moles of $CuS{O}_4$ required to consume 0.01 mol of phosphine (Q) =
$(\frac{3}{2}\times 0.01)\text{mol}$
Molar mass of ${\text{CuSO}}_4=159\text{g}/\text{mol}$
Mass of ${\text{CuSO}}_4=(\frac{3}{2}\times 0.01)\times 159=2.385\text{g}$