Question

$0.250g$ of an element $M$ reacts with excess fluorine to produce $0.547g$ of the hexafluoride, $M{F}_6$. What is the element?

[Atomic mass of $\text{Cr}=52g$, $\text{Mo}=96g$, $\text{S}=32g$, $\text{Te}=127.6g$]

• A. $\text{Cr}$
• B. $\text{Mo}$
• C. $\text{S}$
• D. $\text{Te}$

Answer 1

The correct option is B $\text{Mo}$

$M+3F_2\to M{F}_6$

Let $m$ be the molar mass of the element.

The molar mass of $M{F}_6$ will be $m+6\times 18.9=m+113.4$ g/mol.

Thus, $m$ g of the element $M$ will give $m+113.4$ g of $M{F}_6$.

Thus, $0.250$ g of the element $M$ will give $\frac{0.250}{m}\times (m+113.4)$ g of $M{F}_6$ which is equal to $0.547$ g.

Hence, $\frac{0.250}{m}\times (m+113.4)=0.547$

$0.250m+28.35=0.547m$

$m=95.5$ g/mol.

Hence, the element is $Mo$ (with molar mass $96$ g/mol).

Hence, the correct option is $B$