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 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.
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)`