Mass concentration of water vapor at 273 K from partial pressure

[Nimmy]

Partial Pressure Help---Please Check

The Partial Pressure of water vapor at T = 273 K is 6.11 mbar. Find the corresponding mass concentration of water vapor. Assume P = 1 atm.


So...

ρ^H20 = (P_H20*M_H20/(R*T)
ρ^H20= (6.11*10^2 Pa)*(18.01 g/mol)/(8.314*273 K)
ρ^H20 = 11004.11/2269.722 = 4.85 g/mol

Correct?

 

[Redbelly98]

Essentially correct, yes -- except for the units. You can either (1) think about what the units associated with "mass concentration" should be, or (2) include the units in R=8.314___?, and work out the units in the calculation. I recommend trying both approaches, in the order given here.