MC24 - Gas Laws + Gas Density and Molar Mass Resources « previous | 10 of 14 | n

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MC24 - Gas Laws + Gas Density and Molar Mass Resources « previous | 10 of 14 | next >> + Gas Density and Molar Mass Pressure and temperature affect the amount of space between gas molecules, which affects the volume and, therefore, the density of the gas since density = volume The molar mass of a substance, however, is a constant and can be used to identify an unknown gas sample. Molar mass is found by dividing the mass of a sample (in grams) by the number of moles in that sample. The number of moles of gas can be calculated using the ideal gas law PV = nRT Part A Calculate the density of oxygen, O2, under each of the following conditions: • STP • 1.00 atn and 35.0 °C Express your answers numerically in grams per liter. Enter the density at STP first and separate your answers by a comma. · Hints O E which can be rearranged as n = PV N = RT density at STP, density at 1 atm and 35.0 °C = g/L Submit My Answers Give Up Given the number of moles of a gas and its molar mass, you can calculate the mass of the gas. Since density is equal to the ratio of the mass and volume, you can then divide by the volume to find density. Alternatively, you can use the ratio n/V from the ideal gas equation where n is the number of moles and V is the volume, and convert from moles per unit volume to grams per unit volume using molar mass Part B To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 2.3-L bulb, then filled it with the gas at 1.50 atm and 23.0 °C and weighed it again. The difference in mass was 3.9 g . Identify the gas. Express your answer as a chemical formula. Hints -- A2+ 6 + 0 ? Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

Part A

i) STP

PV = nRT

n/V = P/RT

= 1atm /0.082057(L atm /mol K) × 273.25K

= 0.0446M

molar mass of O2 = 32g/mol

mass of O2 in 1 Liter = 32g/mol × 0.0446mol =1.4272g

density of O2 = 1.427g/L

ii) at 1atm and 35

n/V = 1atm/0.082057(L atm/mol K) × 308.15K

= 0.0395M

mass of O2 in 1 L = 0.0395mol ×32g/mol = 1.264g

Density of O2 = 1.264g/L

PART B

Answer

N2

Explanation

no of mole , n = PV/RT

= 1.5atm × 2.3L /0.082057(L atm/mol K) × 296.15K

= 0.14197

Molar mass of the gas = 3.9g/0.14197mol = 27.47g/mol

Molar mass indicate the gas is Nitrogen (N2)

  

  

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