7. what is the total pressure of the following mixture of gases in a 20.0 l container at 298 k: 12.0 g he,
4.00 mol h2 and 6.25 atm ne?

P(total) = 14.81 atm

Explanation:

According to the Dalton law of partial pressure,

The pressure exerted by mixture of gases are equal to the sum of partial pressure of individual gases.

P(total) = P1 + P2 + P3+ + Pn

Given data:

Volume of container = 20.0 L

Temperature = 298 K

Mass of He = 12.0 g

Moles of hydrogen = 4.00 mol

Pressure of Ne = 6.25 atm

Solution:

Moles of helium:

Number of moles = mass/molar mass

Number of moles = 12 g/ 4 g/mol

Number of moles = 3 mol

Pressure of helium:

PV = nRT

P = nRT/ V

P = 3 mol × 0.0821 atm.L. mol⁻¹.k⁻¹ × 298 K / 20 L

P = 73.4 atm / 20

p = 3.67 atm

Pressure of hydrogen:

PV = nRT

P = nRT/ V

P = 4 mol × 0.0821 atm.L. mol⁻¹.k⁻¹ × 298 K / 20 L

P = 97.86 atm / 20

P = 4.89 atm

Total pressure:

P(total) = P(He) + P(Ne) + P(H₂)

P(total) = 3.67 atm + 6.25 atm + 4.89 atm

P(total) = 14.81 atm

i think that the answer is c