Question

How many grams of He are necessary to fill a balloon having a volume of 4.50 × 10^{3} mL to a pressure of 1.14 × 10^{3} torr at 25.0ºC?

Answer 1

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.

How do we calculate the grams from moles?

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

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