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?

#### Solution 1

Given,

*p* = 0.1 bar

*V* = 34.05 mL = 34.05 × 10^{–3} L = 34.05 × 10^{–3} dm^{3}

R = 0.083 bar dm^{3} 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}.

#### Solution 2

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

`(P_1V_1)/T_1 = (P_2V_2)/T_2` i.e `(1xx34.05)/(546+ 273) = (1xxV_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 gmol^(-1)`