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**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?