Question

\[{\rm{CHC}}{{\rm{l}}_{\rm{3}}}\] and HF lead to the formation of a compound of fluorine of molecular weight 70. The compound is:
A. Fluoroform
B. Fluorine monoxide
C. Fluorine dioxide
D. Fluoromethanol

Answer 1

Hint: We know that the molecular weight is the summation of masses of all atoms of the molecule. Here, to find the compound, we have to calculate the molecular weight of each of the given compounds.

Complete Step by Step Solution:
The compound given in option B is Fluorine monoxide. Chemical symbol of Fluorine monoxide is \[{\rm{O}}{{\rm{F}}_2}\] . Now, we have to find out the molecular weight of \[{\rm{O}}{{\rm{F}}_2}\]. Atomic masses of oxygen and Fluorine are 16 u, 19 u respectively.
Molecular weight of \[{\rm{O}}{{\rm{F}}_2}\]=\[16 + 3 \times 19 = 16 + 57 = 73\,{\rm{u}}\]
Therefore, option B is not correct.

Option C is fluorine dioxide. The chemical formula is \[{{\rm{O}}_2}{{\rm{F}}_2}\]. Atomic masses of oxygen and Fluorine are 16 u, 19 u respectively.
Therefore, molecular weight of \[{{\rm{O}}_2}{{\rm{F}}_2}\]=\[16 \times 2 + 3 \times 19 = 32 + 57 = 89\,{\rm{u}}\]
Therefore, option C is not correct,

Option D is fluoromethanol (\[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{FO}}\]).
Molecular weight of \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{FO = 12 + 3}} \times {\rm{1 + 19 + 16 = 12 + 3 + 19 + 16 = 50u}}\]
Therefore, option D is not correct.

The compound given in option A is Fluoroform. The chemical formula of fluoroform is \[{\rm{CH}}{{\rm{F}}_{\rm{3}}}\] . Now, we have to calculate the molecular weight of \[{\rm{CH}}{{\rm{F}}_{\rm{3}}}\]. Atomic mass of Carbon, Hydrogen and Fluorine are 12 u, 1 u, 19 u respectively.
Therefore, molecular weight of \[{\rm{CH}}{{\rm{F}}_{\rm{3}}}\]=\[12 + 1 + 3 \times 19 = 12 + 1 + 57 = 70\,{\rm{u}}\]
So, when \[{\rm{CHC}}{{\rm{l}}_{\rm{3}}}\]and HF undergoes a reaction, the compound formed is fluoroform. And the chemical reaction is,
\[{\rm{CHC}}{{\rm{l}}_{\rm{3}}} + {\rm{HF}} \to {\rm{CH}}{{\rm{F}}_3} + 3{\rm{HCl}}\]

Hence, option A is right.

Note: Fluoroform is an important reagent in organic reaction. It belongs to the category of greenhouse gas but it does not deplete ozone layer. It is also non-toxic and inexpensive gas. Fluoroform is soluble in water as both fluoroform and water are of polar nature.