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".