Question

A 1. 07 g sample of a noble gas occupies a volume of 363 ml at 35°c and 678 mmhg. Identify the noble gas in this sample. (r = 0. 08206 l×atm/k×mol)

Answer 1

The identity of the noble gas is the sample is Krypton

Ideal Gas law

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

Learn more on Ideal Gas law here: brainly.com/question/20212888

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