Asample 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?

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

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml