What volume will o.128 g of propane, c3h8 occupy at a pressure golf 485 mm Hg and a temperature of 30.0 c

113 mL

Explanation:

Let's apply the Ideal Gases Law for this propose:

P . V = n . R .T → (n. R . T) / P = V

We need to convert the T° to Absolute T° and pressure from mmHg to atm

T° K = 30°C + 273 = 303K

485 mmHg . 1atm / 760 mmHg = 0.638 atm

Let's replace the information obtained:

V = (n . 0.082 . 303K) / 0.638 atm

n = number of moles → 0.128 g . 1mol / 44g = 0.00291 moles

V = (0.00291 moles . 0.082 . 303K) / 0.638 atm → 0.113 L

The value can be written as 113 mL

1 is the answer to this problem