(06.06 MC) What is the density of aluminum chloride vapor (AlCl3) at 498 K and 0.939 atm? (3 points)

a

1.24 g/L

b

3.06 g/L

c

5.17 g/L

d

7.42 g/L

b- 3.06 g/L

Explanation:

The density (d) is defined as the mass per volume : d = m/V

We can deduce the density expression for any gas from the ideal gas equation:

PV= nRT

n/V=P/RT

Since the number of moles (n) is equal to the mass (m) divided into the molar mass (M) of the gas, we can rearrange the equation as follows:

(m/M)/V = P/RT

So, the density (d) is calculated as:

d= m/V = PM/RT

The molar mass of AlCl₃ is calculated from the molar mass of Al and Cl atoms:

M = (26.9 g/mol Al) + (3 x 35.4 g/mol) = 133.1 g/mol

And we have the data: P= 0.939 atm , T = 498 K, R= 0.082 L.atm/K.mol (because R is a constant)

So, we introduce the data to calculate the density in g/L:

d = PM/RT = (0.939 atm x 133.1 g/mol)/(0.082 L.atm/K.mol x 498 K) = 3.06 g/L

Therefore, the correct answer is b 3.06 g/L

2b2t

explanation:

2b2t