We're asked to find the mass, in #"g"#, of a sample of hydrogen gas containing #3.5 xx 10^30# hydrogen molecules.
To convert from particles to moles, we can use Avogadro's number, #6.022xx10^23"particles"/"mol"#.
After this conversion, we can use the molar mass of #"H"_2#, which is
#overbrace(2)^("two atoms of H per molecule") xx overbrace(1.01"g"/"mol")^"molar mass of H" = color(red)(2.02"g"/"mol"#
to convert from moles to grams.
(The number #1.01"g"/"mol"# is the same number as the relative atomic mass of hydrogen, #1.01# #"amu"#, which on most periodic tables can be found directly beneath the element's symbol.)
Converting from molecules to moles, we have
#3.5 xx 10^30cancel("molecules H"_2)((1"mol H"_2)/(6.022xx10^23cancel("molecules H"_2)))#
# = color(blue)(5.8xx10^6# #color(blue)("mol H"_2#
Now, using the molar mass, let's convert from moles to grams:
#color(red)(5.8xx10^6# #cancel(color(red)("mol H"_2))((2.02"g H"_2)/(1cancel("mol H"_2)))#
#= color(purple)(1.2xx10^7# #color(purple)("g H"_2#
rounded to #2# significant figures, the amount given in the problem.
Therefore, #3.5xx10^30# molecules of #"H"_2# has a mass of #color(purple)(1.2 xx 10^7# grams.
