What we're looking for here is the gas sample's molar mass given its mass, pressure, volume, and temperature. Recalling the gas law, we have
[tex] PV = nRT [/tex] or
[tex] n = \frac{PV}{RT} [/tex]
where R is 0.08206 L atm / mol K, P is the given pressure, T is the temperature, and V is the volume.
Before applying the values given, it is important to make sure that they are to be converted to have consistent units with that of R.
Thus, we have
P = 736/ 729 = 0.968 atm
T = 28 + 273.15 = 301.15 K
V = 250/1000 = 0.250 L
Now, applying these converted values into the gas law, we have
[tex] n = \frac{(0.968 atm)(0.250 L)}{(0.08206 L.atm/mol.K)(301.15 K)} [/tex]
[tex] n = 0.00979 moles [/tex]
Given that the mass of the sample is 0.430 g, we have
[tex] molar mass = \frac{mass}{number of moles} [/tex]
[tex] molar mass = \frac{0.430}{0.00979} = 43.9 [/tex]
Thus, the gas sample has a molar mass of 43.9 g/mol.
