Question

The effect of chlorofluorocarbons (such as \({\rm{CC}}{{\rm{l}}_2}\;{{\rm{F}}_2}\) ) on the depletion of the ozone layer is well known. The use of substitutes, such as \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}\;{\rm{F}}(g)\), for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by \(10.0\;{\rm{g}}\)of each of these compounds at STP:

(a) \({\rm{CC}}{{\rm{l}}_2}\;{{\rm{F}}_2}(g)\)

(b) \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}\;{\rm{F}}(g)\)

Short answer

a)Volume occupied by \(10\;{\rm{g}}\) of \({\rm{CC}}{{\rm{l}}_2}\;{{\rm{F}}_2}\)at \({\rm{STP}} = 1.85\;{\rm{L}}\).

b)Volume occupied by \(10\;{\rm{g}}\)of \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}\;{\rm{F}}\) of at \({\rm{STP}} = 4.66{\rm{L}}\).

Step-by-step solution

Definition of volume

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

Calculate the volume occupied of compound \({\rm{CC}}{{\rm{l}}_2}\;{{\rm{F}}_2}(g)\) at STP

a)

Let us solve the given problem.

We must rearrange\(PV = nRT\)to solve for\(V:V = \frac{{nRT}}{P}\)

If we choose to use\(R = 0.08206\;{\rm{L}}\)\(atm\)\({\rm{mo}}{{\rm{l}}^{ - 1}}\;{{\rm{K}}^{ - 1}}\), then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.

Converting into the right units:

\(\begin{aligned}{}n &= 10\;{\rm{gCC}}{{\rm{l}}_2}\;{{\rm{F}}_2} \times \frac{{1\;{\rm{mol}}}}{{120.91\;{\rm{gCCl}}2\;{\rm{F}}2}}\\ &= 0.0827\;{\rm{mol }}\end{aligned}\)

At STP,

\(\begin{aligned}{}T &= 273\;{\rm{K}},\\P &= 1\;{\rm{atm}}\\V &= \frac{{nRT}}{P}\\ &= \frac{{(0.0827\;{\rm{mol}})\left( {0.08206\;{\rm{L}}\;{\rm{atm}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\;{{\rm{K}}^{ - 1}}} \right)(273\;{\rm{K}})}}{{1\;{\rm{atm}}}}\\& = 1.85\;{\rm{L}}\end{aligned}\)

Therefore, volume occupied by \(10\;{\rm{g}}\) of \({\rm{CC}}{{\rm{l}}_2}\;{{\rm{F}}_2}\)at \({\rm{STP}} = 1.85\;{\rm{L}}\).

Calculate the volume occupied of compound \({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}\;{\rm{F}}(g)\)at STP

b)

Consider the given problem and solve.

Solving for\(V\).

Converting into the right units:

\(\begin{aligned}{} n = 10\;{\rm{gC}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}\;{\rm{F}} \times \frac{{1\;{\rm{mol}}}}{{48.06\;{\rm{gC}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}\;{\rm{F}}}}\\ = 0.208\;{\rm{mol }}\end{aligned}\)

At \({\rm{STP, }}\)

\(\begin{aligned}{}T &= 273\;{\rm{K}},\\P &= 1\;{\rm{atm}}\\V &= \frac{{nRT}}{P}\\& = \frac{{(0.208\;{\rm{mol}})\left( {0.08206\;{\rm{L}}\;{\rm{atm}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\;{{\rm{K}}^{ - 1}}} \right)(273\;{\rm{K}})}}{{1\;{\rm{atm}}}}\\ &= 4.66\;{\rm{L}}\end{aligned}\)

Therefore, volume occupied by\(10\;{\rm{g}}\)of\({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}\;{\rm{F}}\)of at\({\rm{STP}} = 4.66{\rm{L}}\).