# If the vapour density for a gas is #20#, then what is the volume of #"20 g"# of this gas at NTP?

##### 1 Answer

I get about

This is an old term for the ratio of the density to the density of

#rho_V = rho_"gas"/rho_(H_2)#

Densities here are in

#PM = rhoRT# ,where

#M# is molar mass in#"g/mol"# and the remaining variables should be well-known...

#P# is pressure in#"atm"# as long as#R = "0.082057 L"cdot"atm/mol"cdot"K"# .

#T# is temperature in#"K"# .

Thus,

#rho_"gas"/rho_(H_2) ~~ (PM_"gas""/"RT)/(PM_(H_2)"/"RT) ~~ M_("gas")/(M_(H_2))#

Thus, the molar mass of the gas is apparently...

#M_("gas") ~~ overbrace(20)^(rho_V) xx overbrace("2.0158 g/mol")^(M_(H_2)) ~~ "40.316 g/mol"#

And the mols of this would be...

#20 cancel"g gas" xx "1 mol"/(40.316 cancel"g") = "0.4961 mols"#

So, at

#color(blue)(V_(gas)) ~~ (nRT)/P#

#~~ (("0.4961 mols")("0.082057 L"cdot"atm/mol"cdot"K")("293.15 K"))/("1 atm")#

#~~# #color(blue)("11.93 L")#