Agas cylinder contains 1.35 mol he, 2.50 mol ne, and 2.30 mol ar. if the total pressure in the cylinder is 2110 mmhg, what is the partial pressure of each of the components? assume constant temperature.

Explanation:

Given data:

Number of moles of He = n1= 1.35 mol

Number of moles of Ne = n2= 2.50 mol

Number of moles of Ar =n3= 2.30 mol

Total pressure = 2110 mmHg

Partial pressure of each component = ?

Solution:

Total number of moles = 1.35 + 2.50 mol + 2.30 mol

Total number of moles = 6.15 mol

P(He) = [ n₁/ nt ] P(total)

P(He) = [ 1.35 / 6.15 ] × 2110 mmHg

P(He) = [ 0.23 ] × 2110 mmHg

P(He) = 485.3 mmHg

P(Ne) = [ n₂/ nt ] P(total)

P(Ne) = [ 2.50/ 6.15 ]  × 2110 mmHg

P(Ne) = 0.41 × 2110 mmHg

P(Ne) = 865.1 mmHg

P(Ar) = [ n₃/ nt ] P(total)

P(Ar) = [ 2.30/6.15 ]× 2110 mmHg

P(Ar) = [ 0.37 ]× 2110 mmHg

P(Ar) = 780.7 mmHg

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the correct answer is:

a. formation of precipitation