If 2.31 g of the vapor of a volatile liquid is able to fill a 498 ml flask at 100 degrees C and 775 mmHg, how do you calculate the molar mass of the liquid? Also, how do you calculate the density of the vapor under these conditions?

1 Answer
Aug 18, 2016

The molar mass of the gas is 139 g/mol, and its density is 4.64 g/L.

Explanation:

We can use the Ideal Gas Law to solve both of these problems:

#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))" "#

Molar mass

Since #n = "mass"/"molar mass" = m/M#, we can write the Ideal Gas Law as

#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))" "#

We can rearrange this to get

#M = (mRT)/(PV)#

#m = "2.31 g"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "(100 + 273.15) K" = "373.15"#
#P = 775 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "1.020 atm"#
#V = "0.498 L"#

#M = ("2.31 g" × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 373.15 color(red)(cancel(color(black)("K"))))/("1.020" color(red)(cancel(color(black)("atm"))) × 0.498 color(red)(cancel(color(black)("L")))) = "139 g/mol"#

∴ The molar mass is 139 g/mol.

Density

We have seen that .

#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))" "#

We can rearrange this to

#PM = m/VRT#

But #"density"= "mass"/"volume"# or #color(brown)(|bar(ul(color(white)(a/a)color(black)(ρ = m/V)color(white)(a/a)|)))" "#

#PM = ρRT# and

#color(blue)(|bar(ul(color(white)(a/a)ρ = (PM)/(RT)color(white)(a/a)|)))" "#

#P = "1.020 atm"#
#M = "46.01 g/mol"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "(100 + 273.15) K" = "373.15 K"#

#ρ = (1.020 color(red)(cancel(color(black)("atm"))) × "139.3 g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.082 06" color(red)(cancel(color(black)("atm")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 373.15 color(red)(cancel(color(black)("K")))) = "4.64 g/L"#