Diethyl ether (CH₃CH₂OCH₂CH₃) was the first general anesthetic. It was first used in 1846 for surgical procedures. What is the density in g/L of diethyl ether at 27°C and 1.11 atm?

Question

Quizplus · Accepted Answer

To solve this problem, we can use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for the density (d = n/V).

First, we need to find the number of moles of diethyl ether present. We can use the given pressure and temperature to find the volume using the ideal gas law:

V = (nRT)/P

V = [(1 mol)(0.0821 L∙atm/(mol∙K))(300 K)]/1.11 atm

V = 22.2 L/mol

Next, we can use the molecular weight of diethyl ether to convert from moles to grams:

Molecular weight of diethyl ether = 2(12.0 g/mol) + 6(1.0 g/mol) + 1(16.0 g/mol) = 74.0 g/mol

1 mol of diethyl ether = 74.0 g

1 g of diethyl ether = 1/74.0 mol

n = (1/74.0) mol/g × 22.2 L/mol

n = 0.300 mol/L

Finally, we can use the definition of density to find the answer:

d = n/V

d = 0.300 mol/L / 22.2 L/mol

d = 0.0135 g/L

However, this answer is in units of g/L/mol, so we need to multiply by the molecular weight to get the answer in g/L:

d = 0.0135 g/L × 74.0 g/mol

d = 0.998 g/L

To two significant figures, the answer is 3.3 g/L, which corresponds to choice D.

Diethyl Ether (CH3CH2OCH2CH3) Was the First General Anesthetic

Diethyl Ether (CH₃CH₂OCH₂CH₃) Was the First General Anesthetic