Question

what is the volume occupied by 15.2 g of argon gas at a pressure of 1.04 atm and a temperature of 353 k ?

Answer 1

The volume of argon will be 10.57 L.

Apply the ideal gas law, PV = nRT, where n is the number of moles and R is the gas constant.

We must divide the given mass of argon by its molar mass because we have mass and need moles. Argon has a molar mass of 40 g/mol.

So the number of moles of Argon in 15.2g is = $\frac{15.2}{40} = 0.38$mol

Given:

P = 1.04 atm

n = 0.38 mol

R = $0.082057 Latm K^{-1} mol^{-1}$

T = 353 K

Unknown:

V

Solution:

V = $\frac{nRT}{P}$

V = $\frac{0.38*0.082*353}{1.04} = 10.57$ L

To learn more about ideal gas law, here

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