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A sample of hydrogen gas is mixed with water vapor. The mixture has a total pressure of 755 torr and the water vapor has a partial pressure of 24 torr. What amount (in moles) of hydrogen gas is contained in 1.55 L of this mixture at 298 K?

Respuesta :

kenmyna
The moles  of hydrogen that contained in  1.55 L calculated   using  the  ideal gas equation

that is PV =nRT  where n is number of  moles

by making n the subject of formula  n = Pv/RT

P= total pressure - partial pressure of water  vapor = 755-24 = 731  torr
V=1.55 l
T=298 k
R (gas Constant)= 62.364 l.torr/mol.k

n= (731  torr  x 1.55 L)/( 62.364 l.torr/mol.k x298 K ) = 0.06  moles  of hydrogen

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