A sealed, rigid container contains three gases: 28.0 \(\mathrm{g}\) of nitrogen, 40.0 \(\mathrm{g}\) of argon, and 36.0 g of water vapor. If the total pressure exerted by the gases is \(2.0 \mathrm{atm},\) what is the partial pressure of the nitrogen? (A) 0.33 atm (B) 0.40 atm (C) 0.50 \(\mathrm{atm}\) (D) 2.0 \(\mathrm{atm}\)

Short Answer

Expert verified
The partial pressure of nitrogen is 0.5 atm.

Step by step solution

01

Determine the number of moles

First, convert the mass of each gas to moles using their respective molar masses. For nitrogen, Ar(N) = 14, thus molar mass = 28 g/mol. Therefore, number of moles = mass/molar mass = 28.0g / 28 g/mol = 1 mol. Similarly, for argon, Ar(Ar) = 40, thus molar mass = 40 g/mol and number of moles = mass/molar mass = 40.0 g / 40 g/mol = 1 mol. For water vapor, molar mass = 18 g/mol and number of moles = mass/molar mass = 36.0 g / 18 g/mol = 2 mol.
02

Calculate the total number of moles

The total number of moles, \(n_{total}\), is the sum of moles of all three gases = 1 mol N2 + 1 mol Ar + 2 mol H2O = 4 mol.
03

Calculate the partial pressure of nitrogen

Using the formula for partial pressure \(P_{gas} = P_{total} * (n_{gas}/n_{total})\), we substitute the values: \(P_{N2} = 2.0 atm * (1 mol / 4 mol) = 0.5 atm\). Therefore, the partial pressure of nitrogen is 0.5 atm.

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