Question

What is the volume of 300. g of mercury vapor at 822K and 0.900 atm?

Answer 1

112.2L

Step-by-step explanation:

Volume (V) = 300g

Temperature (T) = 822K

Pressure (P) = 0.9atm

using the ideal gas equation;

`PV = nRT\\\\ V = (nRT)/(P)`

Molar gas constant (R) =
`0.0821L.atm/mol.K`

Mole (n) =
`(Mass (m))/(Molar mass (M))` Molar mass of Mercury = 200.59g/mol

`n = (300g)/(200.59 g/mol) \\`

= 1.496mol

Now, the volume can be calculated;

V =
`(1.496mol* 0.0821L.atm/mol.K*822K)/(0.9atm)`

∴Volume of mercury =
`112.2L`