Question

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?

Answer 1

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"`