Question

Noble gases (Group 18 in the periodic table) have the following volume concentrations in dry air: $\mathrm{He},5.24\mathrm{ppm};$$\mathrm{Ne},18.2\mathrm{ppm};\mathrm{Ar},0.93\mathrm{vol}\%;\mathrm{Kr},1.14\mathrm{ppm},\mathrm{Xe},87\mathrm{ppb}.$

(a)A concentration of5.24ppm He means role="math" localid="1667555346685" $2.25\mathrm{\mu L}$of He perliter of air. Using the ideal gas law in Problem $\mathbf{1}\mathbf{-}\mathbf{18}$,find how many moles of Heare contained in role="math" localid="1667555352318" $5.25\mathrm{\mu L}$of $\mathrm{He}$at $25.00°\mathrm{C}\left(298.15\mathrm{K}\right)$and 1.000 bar. This number is the molarity ofin the air.

(b)Find the molar concentrations ofAr,Kr, and Xein air at role="math" localid="1667555013049" $25°\mathrm{C}$and1 bar

Short answer

The molarity of He is$2.12\times {10}^{-7}\mathrm{M}$

The molar concentration of Kr is$4.61\times {10}^{-8}\mathrm{M}$

Step-by-step solution

The molarity of

Volume$=5.24\times {10}^{-6}\mathrm{L}$

Pressure$=1.000\mathrm{bar}=0.99\mathrm{atm}$

Temperature$=298.15\mathrm{K}$

The molarity of

(a)

The formula below can be used to solve the number of moles:

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \mathrm{No}.\mathrm{of}\mathrm{moles}=\frac{\left(\mathrm{P}\right)\left(\mathrm{V}\right)}{\left(\mathrm{R}\right)\left(\mathrm{T}\right)} \\ =\frac{\left(0.99\mathrm{bar}\right)\left(5.24\times {10}^{-6}\mathrm{L}\right)}{\left(0.0821{\displaystyle \frac{\mathrm{atm}·\mathrm{L}}{\mathrm{mol}·\mathrm{K}}}\right)\left(298.15\mathrm{K}\right)} \\ =2.12\times {10}^{-17}\mathrm{moles}. \end{gathered}$$

The molarity of He is:

$$\begin{gathered} \mathrm{No}.\mathrm{of}\mathrm{moles}=2.12\times {10}^{-7} \\ \mathrm{Molarity}\mathrm{of}\mathrm{He}=\frac{2.12\times {10}^{-7}}{1} \\ =2.12\times {10}^{-7}\mathrm{M} \end{gathered}$$

Therefore, the molarity of is$2.12\times {10}^{-7}\mathrm{M}$

The molarity of

(b)

The molar solution of Kr is calculated as shown below:

The volume of$\mathrm{Kr}=1.14\times {10}^{-6}\mathrm{L}$

role="math" localid="1667554429225" $$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \mathrm{No}.\mathrm{of}\mathrm{moles}=\frac{\left(\mathrm{P}\right)\left(\mathrm{V}\right)}{\left(\mathrm{R}\right)\left(\mathrm{T}\right)} \\ =\frac{\left(0.99\mathrm{bar}\right)\left(1.14\times {10}^{-6}\mathrm{L}\right)}{\left(0.0821{\displaystyle \frac{\mathrm{atm}·\mathrm{L}}{\mathrm{mol}·\mathrm{K}}}\right)\left(298.15\mathrm{K}\right)} \\ =4.61\times {10}^{-8}\mathrm{L} \end{gathered}$$

role="math" localid="1667554376156" $$\begin{gathered} \mathrm{Molarity}\mathrm{of}\mathrm{Kr}=\frac{4.61\times {10}^{-8}}{1} \\ =4.61\times {10}^{-8}\mathrm{M} \end{gathered}$$

The molar concentration ofKr is$4.61\times {10}^{-8}\mathrm{M}$