Calculate the density of Ar(g)at -11°C and 675 mmHg.
A)-39.3 g/L
B)1.52g/L
C)1.65 g/L
D)39.95 g/L
E)1254 g/L

Question

Quizplus · Accepted Answer

To solve for density, we need to use the ideal gas law: PV=nRT. We can rearrange this equation to solve for n/V, which is the definition of density:

n/V = P/RT

We are given P (675 mmHg) and T (-11°C), but we need to convert T to Kelvin:

T = -11°C + 273.15 = 262.15 K

We also know R, the ideal gas constant (0.08206 L atm/mol K), and the molar mass of Ar (39.95 g/mol).

Plugging in these values:

n/V = (675 mmHg)/(0.08206 L atm/mol K * 262.15 K) = 32.93 mol/L

To get density, we need to multiply by the molar mass:

density = (32.93 mol/L) * (39.95 g/mol) = 1315.48 g/L

However, this answer is in grams per liter and the choices are in g/L. Dividing by 1000 gives:

density = 1.315 g/L

This is closest to choice C, 1.65 g/L.

Calculate the Density of Ar(g)at -11°C and 675 MmHg