Q5.30P - Chemistry: Molecular Nature Of Matter And Change

Question

You have 357 mL of chlorine trifluoride gas at 699 mmHg and $\mathbf{45}\mathbf{°}\mathit{C}$. What is the mass (in g) of the sample?

Step-by-Step Solution

Answer

Answer

The mass of the gas is 1.165 g.

Step 1: Ideal gas law

The ideal gas law equation is, 

$\mathit{P}\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$

The ideal gas law for the given condition can be written as,

$\mathit{n}\mathbf{=}\frac{\mathbf{PV}}{\mathbf{RT}}$

Here,

V is the volume (375 mL)

T is the temperature ($\mathbf{45}\mathbf{°}\mathit{C}$).

P is the pressure (699 mmHg).

Step 2: Determination of mass

Before putting the values, we need to convert the given units in desired units,

$$\begin{gathered} \mathit{V}\mathbf{=}\mathbf{375}\mathbf{ }\mathit{m}\mathit{L}\mathbf{=}\mathbf{375}\mathbf{ }\mathit{m}\mathit{L}\mathbf{\times }\frac{\mathbf{1}\mathbf{L}}{{\mathbf{10}}^{\mathbf{3}}\mathbf{ }\mathbf{m}\mathbf{L}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{375}\mathbf{ }\mathit{L} \end{gathered}$$

And,

$\mathit{T}\mathbf{=}\mathbf{45}\mathbf{°}\mathit{C}\mathbf{=}\mathbf{45}\mathbf{+}\mathbf{273}\mathbf{=}\mathbf{318}\mathbf{ }\mathit{K}$

And, 

$$\begin{gathered} \mathbf{760}\mathbf{ }\mathit{m}\mathit{m}\mathit{H}\mathit{g}\mathbf{=}\mathbf{1}\mathbf{ }\mathit{a}\mathit{t}\mathit{m} \\ \mathbf{699}\mathbf{ }\mathit{m}\mathit{m}\mathit{H}\mathit{g}\mathbf{=}\mathbf{699}\mathit{m}\mathit{m}\mathit{H}\mathit{g}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{a}\mathbf{t}\mathbf{m}}{\mathbf{760}\mathbf{ }\mathbf{m}\mathbf{m}\mathbf{H}\mathbf{g}} \\ \mathbf{699}\mathbf{ }\mathit{m}\mathit{m}\mathit{H}\mathit{g}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{919}\mathbf{ }\mathit{a}\mathit{t}\mathit{m} \end{gathered}$$

The number of moles of gas in the sample is,

$$\begin{gathered} \mathit{n}\mathbf{=}\frac{\mathbf{PV}}{\mathbf{RT}} \\ \mathit{n}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{919}\mathbf{ }\mathbf{atm}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{375}\mathbf{L}}{\mathbf{0}\mathbf{.}\mathbf{0821}{\mathbf{atmLK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}\mathbf{\times }\mathbf{318}\mathbf{K}} \\ \mathit{n}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0126}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

The molar mass of chlorine trifluoride is $\mathbf{92}\mathbf{.}\mathbf{45}\mathbf{ }\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$.

The mass of the sample is,

$$\begin{gathered} \mathit{M}\mathit{a}\mathit{s}\mathit{s}\mathbf{=}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{\times }\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{ }\mathit{m}\mathit{a}\mathit{s}\mathit{s} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0126}\mathit{m}\mathit{o}\mathit{l}\mathbf{\times }\mathbf{92}\mathbf{.}\mathbf{45}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{165}\mathbf{ }\mathit{g} \end{gathered}$$

Thus, the mass of the gas is 1.165 g.