Question

In Victor Mayer's method 0.2g of a volatile compound on volatilisation gave 56mL of vapour at STP.It's molecular weight is?

Answer 1

The molecular mass of the compound is approximately 80 u.

Explanation:

We can use the Ideal Gas Law to solve this problem:.

`color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "`

Since `n = m/M`, we can rearrange this equation to get

`pV = (m/M)RT`

And we can solve this equation to get

`M = (mRT)/(pV)`

STP is defined as 1 bar and 0 °C.

In your problem,

`m = "0.2 g"`
`R = "0.083 14 bar·L·K"^"-1""mol"^"-1"`
`T = "273.15 K"`
`p = "1 bar"`
`V = "56 mL" = "0.056 L"`

∴ `M = ("0.2 g" × "0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar"))) × 0.056color(red)(cancel(color(black)("L")))) = "80 g/mol"`

∴ The molar mass of the compound is 80 g/mol, so the molecular mass is 80 u.

Note: The answer can have only one significant figure, because that is all you gave for the mass of the compound.