Question

A container contains 30 g gas A (molar mass=60g/mol ) and 90 g gas B (molar mass=45g/mol ). The total pressure is 100 mm of Hg. What would the partial pressure (in mm Hg) of gas A be?

Answer 1

Answer:

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Explanation:

A container contains 30 g gas A (molar mass=60g/mol ) and 90 g gas B (molar mass=45g/mol ). The total pressure is 100 mm of Hg. What would the partial pressure (in mm Hg) of gas A be?

Answer 2

A container contains 30 g gas A (molar mass=60g/mol ) and 90 g gas B (molar mass=45g/mol ). The total pressure is 100 mm of Hg. The partial pressure (in mm Hg) of gas A is 20 mm Hg.

• The partial pressure of a gas is equal to the total pressure multiplied by the mole fraction of the gas. The mole fraction of a gas is equal to the number of moles of the gas divided by the total number of moles of all the gases present.

• To find the number of moles of gas A, we divide the mass of gas A by its molar mass:

• n(A) = 30g / 60g/mol = 0.5 mol

• To find the number of moles of gas B, we divide the mass of gas B by its molar mass:

• n(B) = 90g / 45g/mol = 2 mol

• To find the total number of moles of all the gases, we add the number of moles of gas A and gas B:

• n(total) = 0.5 mol + 2 mol = 2.5 mol

• To find the mole fraction of gas A, we divide the number of moles of gas A by the total number of moles:

• X(A) = 0.5 mol / 2.5 mol = 0.2

• Finally, to find the partial pressure of gas A, we multiply the total pressure by the mole fraction of gas A:

• P(A) = 100 mm Hg * 0.2 = 20 mm Hg

  Therefore, the partial pressure of gas A is 20 mm Hg.