Chemistry - Stoichiometry Question with Solution | TestHub

Explain Question : Molarity of ${\mathrm{FeCl}}_3(\mathrm{aq})$

Concept : Concentration term

Solution : Mole of $\mathrm{Fe}[\mathrm{Cl}{]}_3=$ Mole of $\mathrm{Fe}[\mathrm{OH}{]}_3$

$\frac{2.14}{107}=0.02$ mole

$\mathrm{V}=100\mathrm{mL}$

Molarity $=\left(\frac{\mathrm{w}}{\mathrm{Mw}}\right)\times \frac{1000}{{\mathrm{V}}_{\mathrm{mL}}}=\mathrm{n}\times \frac{1000}{{\mathrm{V}}_{\mathrm{mL}}}$

$=0.02\times \frac{1000}{100}=0.2\mathrm{M}$

Moles of ${\mathrm{FeCl}}_3=$ Moles of $\mathrm{Fe}(\mathrm{OH}{)}_3$

$$=\frac{2.14}{107}=0.02\text{ Final Answer }:\left(C\right)$$

$\because 0.02$ moles are present in 100 ml So in 1000 ml , the moles will be $=\frac{0.02}{100}\times 1000=0.2$