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Jet001 [13]
3 years ago
15

At a certain temperature, ethane has a vapor pressure of 304 torr and propane has a vapor pressure of 27 torr. What is the mole

fraction of ethane in the vapor above a solution containing equal molar amounts of ethane and propane?

Chemistry
1 answer:
SOVA2 [1]3 years ago
8 0

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}

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}

\text{The vapour pressure above the solution is $\boxed{\textbf{166 torr}}$}

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The reaction between potassium metal and water produces potassium hydroxide and hydrogen gas.
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Answer:

2K   +   2H₂O   →   2KOH  + H₂

Explanation:

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    Potassium metal  = K

    Water  = H₂O

 The products are:

  Potassium hydroxide  = KOH

  Hydrogen gas  = H₂  

The reaction equation is given as;  

                Reactants   →   Products

        2K   +   2H₂O   →   2KOH  + H₂

The reaction is a single displacement reaction

7 0
3 years ago
A vessel of 120ml capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vess
Rainbow [258]

Given parameters:

Initial volume  = 120ml

Initial temperature  = 35°C

Initial pressure  = 1.2bar

Final volume  = 180ml

Final temperature  = 35°C

Unknown:

Final pressure  = ?

To solve this problem, we apply the combined gas law. The expression is given below;

          \frac{P_{1}V_{1}  }{T_{1} }  = \frac{P_{2}V_{2}  }{T_{2} }

Where P₁ is the initial pressure

           P₂ is the final pressure

          V₁ is the initial volume

          V₂ is the final volume

          T₁ is the initial temperature

           T₂ is the final temperature

We need to convert the parameters to standard units

take the volume to dm³;

      1000ml  = 1dm³

      120ml  = \frac{120}{1000} dm³  = 0.12dm³ = initial volume

Final volume;

      1000ml = 1dm³

      180ml  = \frac{180}{1000} dm³  = 0.18dm³

Now, the temperature;

       K  = 273 + °C

Initial temperature  = 273 + 35  = 308k

Final temperature  = 308k

We then input the parameters into the equation;

         \frac{1.2bar x 0.12 }{308}   = \frac{P_{2} x 0.18 }{308}

       Solving for P₂;

     P₂  = 0.8bar

The new pressure or final pressure in the vessel is 0.8bar

5 0
3 years ago
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