Chemistry - Concentration Terms Question with Solution | TestHub

(I) ${\mathrm{NH}}_3(\mathrm{g})\to \frac{1}{2}{\mathrm{N}}_2(\mathrm{g})+\frac{3}{2}{\mathrm{H}}_2(\mathrm{g})$

Initially: 1 mole

Finally: $0\frac{1}{2}\frac{3}{2}$

$$\begin{array}{rl}& \mathrm{x}=\frac{2-1}{1}\times 100=100\%\\ & {\mathrm{M}}_2={\mathrm{M}}_{\text{avg }}=\frac{\text{ total mass }}{\text{ total mole }}=\frac{1\times 17}{2}=8.5\\ & {\mathrm{M}}_1=17\end{array}$$

(II) ${\mathrm{SO}}_3(\mathrm{g})\to {\mathrm{SO}}_2(\mathrm{g})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})$

Initially: 1 mole

Finally: $01\frac{1}{2}$

$$\begin{array}{rl}& {\mathrm{M}}_2={\mathrm{M}}_{\text{avg }}=\frac{1\times 80}{1.5}\\ & {\mathrm{M}}_1=80\end{array}$$

(III) ${\mathrm{PCl}}_5(\mathrm{g})\to {\mathrm{PCl}}_3(\mathrm{g})+{\mathrm{Cl}}_2(\mathrm{g})$

$t=01$ mole

$t=t01$ mole 1 mole

${\mathrm{M}}_2={\mathrm{M}}_{\text{avg }}=\frac{1\times 208.5}{2}$

${\mathrm{M}}_1=208.5$

(IV) $\mathrm{HCl}(\mathrm{g})\to \frac{1}{2}{\mathrm{H}}_2(\mathrm{g})+\frac{1}{2}{\mathrm{Cl}}_2(\mathrm{g})$

$t=01$ mole

$t=t0\frac{1}{2}$ mole $\frac{1}{2}$ mole

${\mathrm{M}}_2={\mathrm{M}}_{\text{avg }}=\frac{1\times 36.5}{1}$

${\mathrm{M}}_1=36.5$