Question #428df

1 Answer
Jun 21, 2017

V = 56.0 "L"

Explanation:

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".