Respuesta :
Answer:
Volume of 32.0 g [tex]NO_{2}[/tex] at 3.12 atm and 18.0 [tex]^{0}\textrm{C}[/tex] is 5.33 L.
Explanation:
Molar mass of [tex]NO_{2}[/tex] = 46.0055 g/mol
So, 32.0 g of [tex]NO_{2}[/tex] = [tex]\frac{32.0}{46.0055}[/tex] mol of [tex]NO_{2}[/tex] = 0.696 mol of [tex]NO_{2}[/tex]
Let's assume [tex]NO_{2}[/tex] behaves ideally.
Then [tex]P_{NO_{2}}V_{NO_{2}}=n_{NO_{2}}RT[/tex] , where P, V, n, R and T represents pressure, volume, number of moles, gas constant and temperature (in kelvin scale) respectively.
Here [tex]P_{NO_{2}}[/tex] = 3.12 atm, [tex]n_{NO_{2}}[/tex] = 0.696 mol, R = 0.0821 L.atm/(mol.K) and T = (273+18.0)K = 291 K
So, [tex]V_{NO_{2}}=\frac{n_{NO_{2}}RT}{P_{NO_{2}}}[/tex]
or, [tex]V_{NO_{2}}=\frac{(0.696mol)\times (0.0821\frac{L.atm}{mol.K})\times (291K)}{3.12atm}[/tex]
or, [tex]V_{NO_{2}}=5.33L[/tex]
So, volume of 32.0 g [tex]NO_{2}[/tex] at 3.12 atm and 18.0 [tex]^{0}\textrm{C}[/tex] is 5.33 L.
The Volume of NO₂ is 542.8L
What is Ideal Gas Law ?
Ideal Gas Law relates the macroscopic property of gas , It states that the product of pressure and volume of 1 mole of a gas is equal to the product of absolute temperature and universal gas constant.
For n mole of gas,
PV = nRT
In the question it is given that mass of NO₂ is 32 g
P = 3.12 atm
T = 18°Celsius = 291 K
Volume = ?
n = m/M = 32/46 = 0.7
3.12 * V = 0.7*0.0821*291
V = 5.36 L
Therefore the Volume of NO₂ is 5.36L
To know more about Ideal Gas
https://brainly.com/question/8711877
#SPJ5
