A mixture of 0.20 mole of Ar, 0.20 mole of N2 and 0.30 mole of He exerts a total pressure of 2.1 atm. The partial pressure of He in the mixture is?

Assessment: JAMB UTME - Chemistry - 1997 Subject: Chemistry

Question 1 Report

A mixture of 0.20 mole of Ar, 0.20 mole of N2 and 0.30 mole of He exerts a total pressure of 2.1 atm. The partial pressure of He in the mixture is?
Answer Details
We are given a mixture of Ar, N2, and He gases with their respective mole amounts and total pressure. To find the partial pressure of He, we can use Dalton's Law of Partial Pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases. First, we need to calculate the total number of moles of gas in the mixture: 0.20 mol Ar + 0.20 mol N2 + 0.30 mol He = 0.70 mol total Next, we can use the following formula to calculate the partial pressure of He: partial pressure of He = (moles of He / total moles of gas) x total pressure partial pressure of He = (0.30 mol / 0.70 mol) x 2.1 atm partial pressure of He = 0.90 atm Therefore, the partial pressure of He in the mixture is 0.90 atm. Answer choice (a) is correct.

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