Question

34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

Answer 1

Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × 10^–3 L = 34.05 × 10^–3 dm³

R = 0.083 bar dm³ K^–1 mol^–1

T = 546°C = (546 + 273) K = 819 K

The number of moles (n) can be calculated using the ideal gas equation as:

`"pV" = "nRT"`

`=> "n" = ("pV")/("RT")`

`= (0.1 xx 34.05 xx 10^(-3))/(0.083 xx 819)`

`= 5.01 xx 10^(-5)`  mol

Therefore, molar mass of phosphorus = `(0.0625)/(5.01 xx 10^(-5)) = 1247.5 " g mol"^(-1)`

Hence, the molar mass of phosphorus is 1247.5 g mol^–1.

Answer 2

Calculation of volume at `0^@" C"` and 1 bar pressure

`("P"_1"V"_1)/"T"_1 = ("P"_2"V"_2)/"T"_2`   i.e

`(1xx34.05)/(546 + 273) = (1xx "V"_2)/273 or  "V"_2 = 11.35 " ml"`

11.35 mL of vapour at 0°C and 1 bar pressure weight = 0.0625 g

∴ 22700 mL of vapour at 0°C and 1 bar pressure will weigh

`= 0.0625/11.35 xx 22700 = 125" g"`

∴ Molar mass = `125 " g mol"^(-1)`

Alternatively using 

R  = 0.083 bar `" dm"^3 "K"^(-1) "mol"^(-1)`

PV = nRT, i.e

`"n"= ("PV")/("RT") - (1.0 " bar" xx(34.05 xx 10^(-3) " dm"^(3)))/(0.003 " bar" " dm"^(3) "K"^(-1) "Mol"^(-1) xx 819 "K")` 

`= 5 xx 10^(-4) " mol"`

∴ Mass of 1 mole = `0.0625/(5 xx 10^(-4))" g" = 125 " g"`

∴ `"Molar mass"  = 125 " g mol"^(-1)`