Respuesta :

Answer:

0.908 L.

Explanation:

  • Firstly, we should calculate the number of moles of 1.60 g argon.
  • n of Ar = mass / molar mass of Ar = (1.60 g) / (39.948 g/mol) = 0.04 mol.
  • |t is established  and known that the molar volume of a gas at STP (Standard Temperature and Pressure) is equal to 22.4 L for 1 mole of any ideal gas at a temperature of 273.15 K and a pressure of 1.00 atm.

using cross multiplication:

1.0 mole of Ar gas will occupy → 22.7 L

0.04 mole of Ar gas will occupy → ??? L

∴ The volume  occupied by Ar = (0.04 mole) (22.7 L) / (1.0 mole) = 0.908 L.


Answer:

The correct answer is 0.89 L

Explanation:

To find the number of moles,

Moles =  n=m/M.

n = number of moles, m is the mass in grams, and M is the molecular weight in grams per mole.

Given ,

Mass= 1.60g

Molar mass of Argon = 39.94g

[tex]\frac{1.60}{39.94}[/tex]

⇔ 0.04 mole

Now, 1 mole of gas takes up 22.4 L of volume.

0.04 mole = 22.4 × 0.04 =0.89 L

Thus, the volume occupied by the Argon is 0.89L

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