The identity of the noble gas is the sample is Krypton
From the question, we are to determine the identity of the noble gas in the sample
From the ideal gas equation, we have that
PV = nRT
∴ n = PV / RT
Where P is the pressure
V is the volume
n is the number of moles
R is the gas constant
and T is the temperature
From the given information,
P = 678 mmHg = 0.892105 atm
V = 363 mL = 0.363 L
R = 0.08206 L.atm/mol.K
T = 35 °C = 35 + 273.15 K = 308.15 K
Putting the parameters into the equation, we get
n = (0.892105 × 0.363)/ (0.08206 × 308.15)
n = 0.0128 moles
Now, we will determine the Atomic mass of the sample
Using the formula,
Atomic = Mass / Number of moles
Atomic mass of the substance = 1.07 / 0.0128
Atomic mass of the substance = 83.6 amu
The noble gas with the closest atomic mass to this value is Krypton.
Molar mass of Krypton = 83.798 amu
Hence, the identity of the noble gas is the sample is Krypton
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