Question

In an evacuated closed isolated chamber at ${250}^{o}C$. $.02$ mole $P{Cl}_5$ and $.01$ mole ${Cl}_2$ are mixed $(P{Cl}_5\rightleftharpoons P{Cl}_3+{Cl}_2)$. At equilibrium density of mixture was $2.49g/L$ and pressure was $1$ atm. The number of total moles at equilibrium will be approximately:

• A. $0.012$
• B. $0.022$
• C. $0.032$
• D. $0.046$

Answer 1

The correct option is D $0.046$

$\because PV=\frac{m}{M}RT$

$\Rightarrow PM=dRT$

$\therefore M_{mix}=\frac{d_{mix}RT}{P}=\frac{2.49\times 0.0821\times 523}{1}=106.91g$

$M_{mix}=x_{PC{l}_5}{M}_{PC{l}_5}+x_{C{l}_2}{M}_{C{l}_2}+x_{PC{l}_3}{M}_{C{l}_3}$

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

$0.02-\alpha \rightleftharpoons \alpha +0.01+\alpha$

Total moles$=0.02-\alpha +\alpha +0.01+\alpha =0.03+\alpha$

$\therefore 106.91=\frac{(0.02-\alpha )208.5}{0.03+\alpha }+\frac{(0.01+\alpha )71}{0.03+\alpha }+\frac{\alpha \times 137.5}{0.03+\alpha }$

Solving this, we get $\alpha =0.0157$

$\therefore$ Total moles$=0.03+0.0157=0.0457=0.046moles$