(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

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i