Ask Question
25 November, 23:16

A 3.0 L cointainer contains 4 mol He, 2 mol Ne, and 1 mol Ar. What is the mole fraction of neon gas? What is the partial pressure of Ne if the total pressure is 5 atm

+1
Answers (2)
  1. 25 November, 23:18
    0
    Partial pressure of Ne = 1.43 atm

    Explanation:

    Step 1: Data given

    Volume = 3.0 L

    Number of moles He = 4 moles

    Number of moles Ne = 2 moles

    Number of moles Ar = 1 mol

    Total pressure = 5 atm

    Step 2: Calculate total moles

    Total moles = 4 moles + 2 moles + 1 mol

    Total moles = 7 moles

    Step 3: Calculate the mol ratio

    Mol ratio = moles / total mol

    Mol ratio He = 4 moles / 7 moles = 0.571

    Mol ratio Ne = 2 moles / 7 moles = 0.286

    Mol ratio Ar = 1 mol / 7 moles = 0.143 moles

    Ste p4: Calculate partial pressure

    PArtial pressure = mol ratio * total pressure

    Partial pressure of Ne = mol ratio Ne * total pressure

    Partial pressure of Ne = 0.286 * 5 atm

    Partial pressure of Ne = 1.43 atm
  2. 25 November, 23:20
    0
    1. Mole fraction of Ne is 2/7

    2. The partial pressure of Ne is 1.43 atm

    Explanation:

    Step 1:

    Data obtained from the question. This includes:

    Mole of He = 4 moles

    Mole of Ne = 2 moles

    Mole of Ar = 1 mole

    Total mole = 7 moles

    Total pressure = 5 atm

    Step 2:

    Determination of the mole fraction of Ne.

    Mole fraction of a gas is simply the ratio of the mole of the gas to the total mole of the gas present.

    Mole fraction of Ne = mol of Ne / total mole

    Mole fraction of Ne = 2/7

    Step 3:

    Determination of the partial pressure of Ne.

    Partial pressure = mole fraction x total pressure

    Partial pressure of Ne = 2/7 x 5

    Partial pressure of Ne = 1.43 atm
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A 3.0 L cointainer contains 4 mol He, 2 mol Ne, and 1 mol Ar. What is the mole fraction of neon gas? What is the partial pressure of Ne if ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers