The mass in grams of He are necessary to fill a balloon having a volume of 4.50×10³ mL to a pressure of 1.14×10³ torr at 25.0ºC is 1.104 grams.
<h3>How do we calculate the grams from moles?</h3>
Mass (W) in grams from moles (n) will be calculated by using the below equation:
n = W/M, where
M = molar mass
Moles of helium gas will be calculated by using the ideal gas equation:
PV = nRT, where
R = universal gas constant = 62.363 L.torr / K.mol
P = pressure of gas = 1.14×10³ torr
V = volume of gas = 4.50×10³ mL = 4.50 L
n = moles of gas = ?
T = temperature = 25.0ºC = 298 K
On putting these values on the ideal gas equation, we get
n = (1140)(4.5) / (62.363)(298) = 5130 / 18584
n = 0.276 moles
Now we convert this moles into mass by using the first equation as:
W = (0.276mol)(4g/mol) = 1.104 g
Hence required mass of helium gas is 1.104 grams.
To know more about ideal gas equation, visit the below link:
brainly.com/question/12873752
#SPJ1
