Answer:
8.22 g/dm³
Explanation:
Applying,
PV = nRT............... Equation 1
Where P = pressure of krypton, V = Volume of krypton, n = number of moles, R = molar gas constant, T = Temperature
But,
number of mole (n) = mass(m)/molar mass(m')
n = m/m'.............. Equation 2
Substitute equation 2 into equation 1
PV = mRT/m'
P = (m/V)(RT)/m'................ Equation 3
Also,
Density (D) = Mass(m)/Volume(V)
D = m/V.............. Equation 4
Substitute equation 4 into equation 3
P = DRT/m'
D = Pm'/RT.................... Equation 5
From the question,
Given: P = 3.00 atm, T = 100°C = (273+100) = 373 K, m' = 83.8 g/mol
Constant: R = 0.082atm.dm³.K⁻¹.mol⁻¹
Substitute these values into equation 5
D = (3×83.8)/(0.082×373)
D = 251.4/30.586
D = 8.22 g/dm³
The density of krypton gas is 8.22 g/dm³
Density of any substance is its mass per unit volume.
By ideal gas law
PV = nRT
Where p = pressure is 3.00 atm,
t = temperature is 100 °C = 273 + 100 = 373 K
Molar mass = 83.8 g/mol
R = constant is 0.082 atm.
[tex]\rm Number\;of \;moles= \dfrac{mass}{molar\;mass}[/tex]
Substituting the equation in 1
P = (m/V)(RT)/m'
Now, density = m/V
Then,
[tex]\rm D = \dfrac{Pm'}{RT}[/tex]
[tex]\rm D = \dfrac{3\times 83.8}{0.082\times373} = 8.22 g/dm^3\\[/tex]
Thus, the density is 8.22 g/dm3.
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