22 April, 21:48

# A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne in the flask. What is the total pressure?

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1. 22 April, 22:48
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pHe = 3.2 * 10⁻³ atm

pNe = 2.5 * 10⁻³ atm

P = 5.7 * 10⁻³ atm

Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 * 10⁻³ g

mNe = 2.05 mg = 2.05 * 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

0.52 * 10⁻³ g * (1 mol / 4.00 g) = 1.3 * 10⁻⁴ mol

We can find the partial pressure of He using the ideal gas equation.

P * V = n * R * T

P * 1.00 L = 1.3 * 10⁻⁴ mol * (0.082 atm. L/mol. K) * 298 K

P = 3.2 * 10⁻³ atm

The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 * 10⁻³ g * (1 mol / 20.18 g) = 1.02 * 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P * V = n * R * T

P * 1.00 L = 1.02 * 10⁻⁴ mol * (0.082 atm. L/mol. K) * 298 K

P = 2.5 * 10⁻³ atm

The total pressure is the sum of the partial pressures.

P = 3.2 * 10⁻³ atm + 2.5 * 10⁻³ atm = 5.7 * 10⁻³ atm