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A rigid container contains 2.50 mol O2, 0.50 mol N2, and 1.00 mol CO2. The total pressure of the system is 200 kPa. The partial pressure exerted by the N2 in the mixture is Question 10 options:

Question

A rigid container contains 2.50 mol O2, 0.50 mol N2, and 1.00 mol CO2. The total pressure of the system is 200 kPa. The partial pressure exerted by the N2 in the mixture is

Question 10 options:

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Solution

To find the partial pressure exerted by N2, we first need to find the total number of moles in the mixture.

Step 1: Add up all the moles Total moles = moles of O2 + moles of N2 + moles of CO2 Total moles = 2.50 mol + 0.50 mol + 1.00 mol = 4.00 mol

Step 2: Calculate the mole fraction of N2 Mole fraction of N2 = moles of N2 / total moles Mole fraction of N2 = 0.50 mol / 4.00 mol = 0.125

Step 3: Calculate the partial pressure of N2 Partial pressure of N2 = mole fraction of N2 * total pressure Partial pressure of N2 = 0.125 * 200 kPa = 25 kPa

So, the partial pressure exerted by the N2 in the mixture is 25 kPa.

This problem has been solved

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