To solve this problem, we can first convert the given mass of #"He"# to moles, using its molar mass (#4.00"g"/"mol"#, as seen from a periodic table):
#10.0cancel("g He")((1color(white)(l)"mol He")/(4.00cancel("g He"))) = 2.50# #"mol He"#
At standard temperature and pressure conditions (#1# bar and #0^"o""C"#), one mole of an ideal gas has a volume of #22.41# #"L"#.
Therefore, using dimensional analysis again,
#2.50cancel("mol He")((22.41color(white)(l)"L")/(1cancel("mol He"))) = color(red)(56.0# #color(red)("L"#
Thus, #10.0# grams of helium occupies a volume of roughly #color(red)(56.0# #sfcolor(red)("liters"#.
