Let 9102 g He 9535 g F2 and 8846 g Ar be placed in a 129L co

Let 9.102 g He, 9.535 g F2, and 8.846 g Ar be placed in a 12.9-L container at 55°C. What are the molar concentrations and partial pressures of the gases?

Solution

Answer:-

First we calculate the molar concentration

In your case we need the number of moles of He, F2 and Ar then we calculate molar concentration

For He

9.102 g / 4.00 g/mol = 2.2755 mol He

Therefore the formula is,

Molarity = number of moles / volume of solution in liters

Molarity = 2.2755 mole / 12.9 L = 0.176 M

For F2

9.535 g / 37.997 g/mol = 0.2509 mol F2

0.2509 mol / 12.9 L = 0.0195 M

For Ar

8.846g / 39.948g/mol = 0.2214 mol Ar

0.2214 mol / 12.9 L = 0.0172M

We can calculate the partial pressures of each component by means of separate Ideal Gas Equations

P = n RT / V

Where,

P = the pressure

n = the number of moles

R = the universal gas constant

V = the volume

T = the temperature in Kelvin

P(He) = (2.2755 mol × 0.0821 L atm/K mol × 328k) / 12.9 L = 4.75 atm

P(F2) = (0.2509 mol × 0.0821 L atm/K mol × 328k) / 12.9 L = 0.52 atm

P(Ar) = (0.2214 mol × 0.0821 L atm/K mol × 328k) / 12.9 L = 0.46 atm