Question 1(multiple choice worth 3 points)
(06.06 mc)

what is the density of aluminum chloride vapor (alcl3) at 498 k and 0.939 atm?

a.1.24 g/l
b. 3.06 g/l
c. 5.17 g/l
d.7.42 g/l

= 3.064 g/L

Using the equation;

PV = nRT , where P is the pressure,. V is the volume, n is the number of moles and T is the temperature and R is the gas constant, 0.08206 L. atm. mol−1.

Number of moles is 1 since one mole has a mass equivalent to the molar mass.

Therefore; We can find the volume and thus get the density.

V = nRT/P

   = (1 × 0.08206 × 498 )/ 0.939

   = 43.52 L

But; density = mass/volume

Therefore;

Density = 133.34 / 43.52

             = 3.064 g/L