Question #b3b27

Vapor density is simply the density of a gas compared with the density of hydrogen gas, #"H"_2#, kept under the same conditions for pressure and temperature.

In essence, vapor density tells you the ratio that exists between between the mass of a gas present in a given volume and the mass of hydrogen gas present in the same volume.

You can thus say that vapor density is equal to

#color(blue)(|bar(ul(color(white)(a/a)"vapor density" = "molar mass of a gas"/"molar mass of H"_2color(white)(a/a)|)))#

Now, your unknown element is said to have a molar mass of #"27 g mol"^(-1)# and a valency of #3#. A quick look in the periodic table wil reveal that you're dealing with aluminium, #"Al"#.

Aluminium combines with chlorine, #"Cl"#, to form aluminium chloride, #"AlCl"_3#, and ionic compound that contains one aluminium cation, #"Al"^(3+)#, and three chloride anions, #"Cl"^(-)#.

Notice that the cation has a charge of #3+#, which is why the element's valency was said to be equal to #3#.

In order to find the vapor density of aluminium chloride, you must first calculate its molar mass. Another look in the periodic table will show that chlorine has a molar mass of approximately #"35.45 g mol"^(-1)#.

The molar mass of the compound will be

#1 xx "27 g mol"^(-1) + 3 xx "35.45 g mol"^(-1) = "133.35 g mol"^(-1)#

You can take the molar mass of hydrogen gas to be approximately #2 g mol"^(-1)#, which means that the vapor density of the aluminium chloride will be

#"vapor density" = (133.35 color(red)(cancel(color(black)("g mol"^(-1)))))/(2 color(red)(cancel(color(black)("g mol"^(-1))))) = color(green)(|bar(ul(color(white)(a/a)66.675color(white)(a/a)|)))#

SIDE NOTE The actual answer will probably vary a bit depending on the value you pick for the molar mass of chlorine. For example, if you take #"35.5 g mol"^(-1)#, you will end up with

#"vapor density" = 66.75#