Question

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

Answer 1

Answer:

112.2L

Explanation:

Volume (V) = 300g

Temperature (T) = 822K

Pressure (P) = 0.9atm

using the ideal gas equation;

$PV = nRT\\\\ V = \frac{nRT}{P}$

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

Mole (n) = $\frac{Mass (m)}{Molar mass (M)}$                Molar mass of Mercury  = 200.59g/mol

$n = \frac{300g}{200.59 g/mol}$

   = 1.496mol

Now, the volume can be calculated;

V = $\frac{1.496mol* 0.0821L.atm/mol.K*822K}{0.9atm}$

∴Volume of mercury = $112.2L$