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3 years ago

What volume is occupied by 18.4 g oxygen at 28.0°C and a pressure of 0.998 torr? (round to sig figs)

Chemistry

1 answer:

Alexxandr [17]3 years ago

8 0

Answer:

11.0 dm³

Explanation:

From the question,

Applying

PV= nRT............... Equation 1

Where P = pressure of oxygen gas, V = volume of oxygen gas, n = number of moles of oxygen, R = molar constant, T = Temperature.

make V the subeject of the equation

V = nRT/P............. Equation 2

But,

Number of mole (n) = Mass of oxygen(m)/Molar mass of oxygen(m')

n = m/m'....................... Equation 3

Substitute equation 3 into equation 2

V = mRT/Pm'............. Equation 4

Given: T = 28°C = (28+273) = 301 K, P = 0.998 torr = (0.998×0.00131579) = 1.3132 atm, m = 18.4 g

Constant: R = 0.082 atm.dm³/K.mol, m' = 32 g/mol.

Substitute these values into equation 4

V = (301×18.4×0.082)/(32×1.3132)

V = 454.1488/42.0224

V = 10.81 dm³

V = 11.0 dm³

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