Sprint 10:53 PM * 100% (ii),+ t Gas Density and Molar Mass 12417 Constants Perio

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Sprint 10:53 PM * 100% (ii),+ t Gas Density and Molar Mass 12417 Constants Periodic Table Pressure and temperature affect the amount of space between gas molecules, which affects the volume and, therefore, the density of the gas since densityvolume 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 which can be rearranged as n- 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 A Calculate the density of oxygen, O2, under each of the following conditions STP 1.00 atm 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 View Available Hint(s) density at STP, density at 1 atm and 35.0 C- Part B To identify a diatomic gas (X2), aresearcher carried out the following experiment: She weighed an empty 4.3-L bulb, then filled it with the gas at 1.80 atm and 29.0 C and weighed it again. The difference in mass was 8.7 g.Identify the gas. Express your answer as a chemical formula. View Available Hint(s) Provide Feedback Back to UB Mobile

Explanation / Answer

Part A)

Molar mass of O2 = 32 g/mol

1 mol of any gas at STP occupies 22.4 L

At STP, density = 32 gmol-1 / 22.4 Lmol-1 = 1.4286 g/L

Again, at 1.00 atm and 35.00C we have,

PV =nRT

=> n/V = P/RT = 1.00 atm / 0.08206 L atm mol-1K-1 x (35.0 +273) K

=> n/V = 1.00 atm / 0.08206 L atm mol-1K-1 x 308 K
=> n/V = (1.00 / 25.27448) mol/L

=> n/V = 0.0395656 mol/L

Now density = 32 gmol-1 x 0.0395656 molL-1 = 1.27 g/L.

Part B)

Given, V = 4.3 L, P = 1.80 atm , T = 29.00C = (29+273) K = 302 K

Now, PV= nRT

=> n = PV/RT

=> n = 1.80 atm x 4.3 L / 0.08206 L atm mol-1K-1 x 302 K

=> n = 7.74 / 24.78212 mol

=> n = 0.3123 mol

Now, molar mass of X2 = m/n = 8.7 g/0.3123 mol = 27.85 g/mol = 28 g/mol (approximately)

Thus, identity of X2 will be nitrogen gas ,N2(g).

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