The density of a certain gaseous fluoride of phosphorus is \(3.93\;{\rm{g}}/{\rm{L}}\)at STP. Calculate the molar mass of this fluoride and determine its molecular formula.

A substance's volume is the amount of space it occupies

Let us solve the given problem.

We have,

\(PV = nRT\)

where,

\(P = \)Pressure of the gas.

\(V = \)volume of the gas.

\(n = \)Number of moles of the gas.

\(T = \)Temperature of the gas.

\(\begin{aligned}{}R& = {\rm{ Ideal gas constant}}\\& = 0.08206\;{\rm{L atm mo}}{{\rm{l}}^{ - 1}}{K^{ - 1}}\end{aligned}\)

If, amount of the gas\((n)\)is in moles, temperature\((T)\)is in Kelvin\((K)\), pressure\((P)\)is in atm.

\(\begin{aligned}{}\frac{P}{{RT}} &= \frac{n}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \frac{{n \cdot (\mu )}}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \frac{m}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \rho \end{aligned}\)

Density\((\rho ) = \frac{{P \cdot (\mu )}}{{RT}}\)

Where\(\mu = \)molar mass.

Consider the given problem.

\({\rm{density }}(\rho ) = 3.93\frac{{{\rm{ grams }}}}{L}.\)

Given that condition are at STP

That is:

\(\begin{aligned}{}{\rm{Volume }}(V) & = 22.4L.{\rm{ Pressure }}\\(P) &= 1\;{\rm{atm}}.{\rm{ Temperature }}\\(T) = 273.15K.{\rm{ Density}}\\{\rm{ }}(\rho ) &= \frac{{P \cdot (\mu )}}{{RT}}\\ &= \frac{{(3.93) \cdot (0.08206) \cdot (273.15)}}{1}\\ &= 88.08\frac{{{\rm{ grams }}}}{{{\rm{ mole }}}}.\end{aligned}\)

Therefore, Molar mass \(88.08\frac{{{\rm{ grams }}}}{{{\rm{ mole }}}}.\)

We have to find Potassium Fluoride with molecular weight of \(88.08{\rm{grams}}\).

Atomic weight of Potassium (P) \( = 39.08\)grams.

Atomic weight of Florine \((F) = 19\) gram.

\(\begin{aligned}{}P + x(F) &= 88.08{\rm{ grams}}{\rm{. }}30.08 + x(19)\\& = 88.08\\x(19) = 88.08 - 30.08x(19) = 58\\x &= \frac{{58}}{{19}}\\x &= 3\end{aligned}\)

Hence, the Potassium fluoride is \({\rm{P}}{{\rm{F}}_3}\).