Question

1.61 gm of anhydrous $\text{ZnS}{\text{O}}_{\text{4}}$ was placed in moist air, after few days its weight was found to be 2.87 gm. what is the molecular formula of hydrated salt?

[$Zn=65$]

• A. $\text{ZnS}{\text{O}}_{\text{4}}.{\text{H}}_{\text{2}}\text{O}$
• B. $\text{ZnS}{\text{O}}_{\text{4}}.3{\text{H}}_{\text{2}}\text{O}$
• C. $\text{ZnS}{\text{O}}_{\text{4}}.7{\text{H}}_{\text{2}}\text{O}$
• D. $\text{ZnS}{\text{O}}_{\text{4}}.10{\text{H}}_{\text{2}}\text{O}$

Answer 1

Answer: (C)

$\text{ZnS}{\text{O}}_{\text{4}}.7{\text{H}}_{\text{2}}\text{O}$

Moles of $ZnS{O}_4=\frac{1.61}{161}=0.01$ $moles$

After few days, weight changed to $2.87$. Number of $H_{2}O$ molecule added be $x$.

So, $0.01=\frac{2.87}{161+18x}$

$⟹1.61+0.18x=2.87$

$⟹x=7$