Verified answer

first we find the pressure of the gas mixture:

since the density of water is about 1 gram / ml,

& mercury has a density about 13.5 grams / ml

13.5 times less mm of mercury, will put out as much pressure as

85 mm of liquid water in the tube.

so

85 mm of water / 13.5 = 6.30 mm of Hg

which tells us that the height of water in the tube puts out 6.3 Torr of the 770 total torr

so

the gas mix has a total pressure of 764 Torr

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find the pressure of "dry" hydrogen gas in the tube

total gas pressure is 764 Torr

according to http://home.comcast.net/~frankhanson2/vapor.htm

there is 18.7 Torr of water vapor @ 21 Celsius

so

764 Torr - 19 torr water =

your first answer

745 Torr of dry H2

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b. The number of moles of hydrogen gas in the tube.

PV = nRT

(745 Torr) (0.0800 Litres) = n (62.36 L -Torr / K-mol) (294 Kelvin)

n = 0.003251 moles of H2

your answer,

rounded to 3 sig figs is

0.00325 moles of H2