a. 5.8 grams of unknown gas occupies 1.11 L at 27°C and 1.4 atm. Determine its molar mass.

Answer 1

PV = nRT

This formula describes the ideal gas law

Where P = pressure (atm)

V = Volume (liters)

n = moles

R = 0.08206 (L•atm)/(mol•K)

T = 27°C = 27°C + 273 = 300 K

Substituting the values we are given we get:

(1.4 atm) × (1.11 L) = n × 0.08206 (L•atm)/(mol•K) × (300 K)

Now we must solve for moles, n :

Rearranging:

(PV)⁄(RT) = n

((1.4 atm) × (1.11 L)) ⁄ ((0.08206 (L•atm)/(mol•K)×(300 K)) = n

0.063124543017304411406288081891299049475993175725079210333 mol = n

Molar Mass = grams / mol = 5.8 g ⁄ 0.06312454301730441140628808189129904947... mol

Molar Mass = 91.9 g/mol (using three significant figures)

Answer 2

Since molar mass is grams/mole, we need to find moles, since we already know the grams (5.8 g).

To find moles, use the ideal gas law PV = nRT and solve for n (mol)

n = PV/RT

n = (1 atm)(1.11 L) / (0.0821 Latm/Kmol)(300K)

Once you solve for n, molar mass = 5.8 g/n moles