Determine the molar mass of a 0.458 gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. show work.
!

9.165 g/mol.

Explanation:

PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the container in L.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas in K.

∵ n = mass/molar mass.

∴ PV = (mass/molar mass)RT

∴ molar mass = (mass/PV)*(RT)

∵ m = 0.458 g, R = 0.082 L.atm/mol.K, P = 0.98 atm, V = 1.20 L, T = 287 K.

∴ molar mass = (mass/PV)*(RT) = (0.458 g/(0.98 atm)(1.20 L))*(0.082 L.atm/mol.K)(287 K) = 9.165 g/mol.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.