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