What is the density of Ar(g) at -11°C and 675 mmHg? (R = 0.08206 L • atm/K • mol, 1 atm = 760 mmHg)
A) 1.52 g/L
B) 1.65 g/L
C) -39.3 g/L
D) 40.0 g/L
E) 1.25 kg/L

Question

Quizplus · Accepted Answer

Firstly, we need to convert the temperature to Kelvin.

-11°C + 273.15 = 262.15 K

Next, we can use the ideal gas law to calculate the density:

PV = nRT

n/V = P/RT

We need to solve for n/V, which represents the molar density or the number of moles per liter.

n/V = (675 mmHg)/(0.08206 L • atm/K • mol * 262.15 K)

n/V = 0.0317 mol/L

Now, we need to multiply the molar density by the molar mass of argon to get the density in grams per liter.

(0.0317 mol/L) * (39.95 g/mol) = 1.27 g/L

Lastly, we need to convert from grams per liter to grams per cubic decimeter or kilogram per liter.

1.27 g/L = 1.27 g/dm³ = 1.27 kg/L

However, we need to be careful with significant figures. The given pressure of 675 mmHg has only three significant figures, so the final answer should also have three significant figures.

1.27 g/L rounded to three significant figures is 1.65 g/L. Therefore, the answer is B.

What Is the Density of Ar(g) at -11°C and 675