Answer:
166 torr
Explanation:
Let’s call ethane Component 1 and propane Component 2.
According to Raoult’s Law,
![p_{1} = \chi_{1}p_{1}^{\circ}\\p_{2} = \chi_{2}p_{2}^{\circ}](https://tex.z-dn.net/?f=p_%7B1%7D%20%3D%20%5Cchi_%7B1%7Dp_%7B1%7D%5E%7B%5Ccirc%7D%5C%5Cp_%7B2%7D%20%3D%20%5Cchi_%7B2%7Dp_%7B2%7D%5E%7B%5Ccirc%7D)
where
p₁ and p₂ are the vapour pressures of the components above the solution
χ₁ and χ₂ are the mole fractions of the components
p₁° and p₂° are the vapour pressures of the pure components.
Data:
p₁° = 304 torr
p₂° = 27 torr
n₁ = n₂
1. Calculate the mole fraction of each component
χ₁ = n₁/(n₁ + n₂)
χ₁ = n₁/n₁ + n₁)
χ₁ = n₁/(2n₁)
χ₁ = ½
χ₁ = 0.0.5
χ₂ = 1- χ₁ = 1- 0.5 = 0.5
2. Calculate the vapour pressure of the mixture
![p_{1} = 0.5 \times \text{304 torr} = \text{ 152 torr}\\p_{2} = 0.5 \times \text{27 torr} = \text{ 13.5 torr}\\p_{\text{tot}} = p_{1} + p_{2} = \text{152 torr + 13.5 torr} = \textbf{166 torr}](https://tex.z-dn.net/?f=p_%7B1%7D%20%3D%200.5%20%5Ctimes%20%5Ctext%7B304%20torr%7D%20%3D%20%5Ctext%7B%20152%20torr%7D%5C%5Cp_%7B2%7D%20%3D%200.5%20%5Ctimes%20%5Ctext%7B27%20torr%7D%20%3D%20%5Ctext%7B%2013.5%20torr%7D%5C%5Cp_%7B%5Ctext%7Btot%7D%7D%20%3D%20p_%7B1%7D%20%2B%20p_%7B2%7D%20%3D%20%5Ctext%7B152%20torr%20%2B%2013.5%20torr%7D%20%3D%20%5Ctextbf%7B166%20torr%7D)
![\text{The vapour pressure above the solution is $\boxed{\textbf{166 torr}}$}](https://tex.z-dn.net/?f=%5Ctext%7BThe%20vapour%20pressure%20above%20the%20solution%20is%20%24%5Cboxed%7B%5Ctextbf%7B166%20torr%7D%7D%24%7D)
Answer:
A charge of -3
Explanation:
If you look on a periodic table you will find that phosphorus is 3 elements away from the nearest noble gas argon. This means that phosphorus would prefer a charge of -3!