Question
In: Chemistry
A mixture if Krypton and Neon gases, at a total pressure of 678mm Hg, contains 5.96...
A mixture if Krypton and Neon gases, at a total pressure of 678mm
Hg, contains 5.96 grams of krypton and 4.42 grams of neon what is
the partial pressure of each gas in the mixture in mm Hg?
Solutions
Expert Solution
First we need to calculate the mole fraction of each gas in the mixture, with molar mass we can determine that:
5.96 g Kr * (1mol/ 83.798 g) = 0.0711 mol Kr
4.42 g Ne *(1mol/ 20.1797 g) = 0.219 mol Ne
Now the mole fraction could be calculated using the following formula:
For Krypton, we have:
And for the neon we have 0,755 for the mole fraction.
The partial pressure of an individual gas component in an ideal gas can be obtained using this expression:
where: xi is the mole fraction of any individual gas component in a gas mixture, pi is the partial pressure of any individual gas component in a mixture and p is the total pression of the gas mixture.
The partial pressure for Krypton will be:
pKr = 0.254*678 mmHg = 166 mmHg
And the partial pressure for Neon will be:
pKr = 0,755*678 mmHg = 512 mmHg
queen_honey_blossom answered 2 years agoRelated Solutions
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