Question

Wally fluoride is an imaginary gaseous

compound with a molar mass of 314.2 g/mol.
(a) What is the density of wollmanium fluoride at 425 K
and 165 torr?

Answer 1

`\rho =1.96(g)/(L)`

Step-by-step explanation:

Hello there!

In this case, since this imaginary gas can be modelled as an ideal gas, we can write:

`PV=nRT`

Which can be written in terms of density and molar mass as shown below:

`(P)/(RT) =(n)/(V) \\\\(P)/(RT) =(m)/(MM*V)\\\\(P*MM)/(RT) =(m)/(V)=\rho`

Thus, by computing the pressure in atmospheres, the resulting density would be:

`\rho = (165/760 atm * 314.2 g/mol)/(0.08206(atm*L)/(mol*K)*425K) \\\\\rho =1.96(g)/(L)`

Best regards!