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it is desired to inflate a baggie with a volume of 927 mm by filling it with argon gas at a pressure of 0.749 atm and a temperature of 297K how many grams of Ar gas are needed? In g?

Answer :

Willchemistry

Hey here!:

Molar mass Argon = 39.948 g/mol

volume in liter = 927 mL / 1000 => 0.927 L

p = 0.749 atm

T = 297 K

R = 0.0821

Therefore:

Mass of argon=   ( p *  V * molar mass Ar ) / ( R * T )

Mass of argon = ( 0.749 * 0.927 * 39.948  ) / ( 0.0821 * 297 )

Mass of argon = 27.7368 / 24.3837

Mass of argon = 1.137 grams

Hope that helps!

pstnonsonjoku

The mass of the gas in gram is obtained as  1.12 g.

We have the following parameters;

Volume of the gas = 927 mL or 0.927 L

Pressure of the gas = 0.749 atm

Temperature of the gas = 297K

Using the ideal gas equation;

PV =nRT

R = molar gas constant = 0.082 atmLK-1mol-1

n = Number of moles of the gas = ?

n = PV/RT

n = 0.749 atm × 0.927 L/0.082 atmLK-1mol-1  ×  297K

n = 0.694/24.354

n = 0.028 moles

Molar mass of Ar = 40 g/mol

Mass = Number of moles × molar mass

Mass = 0.028 moles ×  40 g/mol

Mass = 1.12 g

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