Question

HELP!!!!! How much does a sample of Argon weigh if it occupies 25.4 L at a pressure of 2.45 atm and a temperature of 482 K?

Answer 1

Answer:

62.82 g

Explanation:

From the question,

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

Where P = Pressure, V = Volume, n = number of moles of argon, R = molar gas constant, T = Temperature.

But,

Number of mole (n) = mass (m)/molar mass(m')

n = m/m'................... Equation 2

Substitute equation 2 into equation 1

PV = (m/m')RT.............. Equation 3

From the question, we were asked to find m.

There make m the subject of formula in equation 3

m = PVm'/RT.............. Equation 4

Given: P = 2.45 atm, V = 25.4 L, T = 482 K

Constant: R = 0.082 atm.dm³.K⁻¹.mol⁻¹, m' = 39.9 g/mol

Substitute these values into equation 4

m = (2.45×25.4×39.9)/(0.082×482)

m = 62.82 g.