# Question 02e7c

Jun 20, 2017

$M W = 143$ $\text{amu}$

#### Explanation:

I'll assume these are the measurements when the liquid and vapor are at dynamic equilibrium.

To solve this problem, we can first use the ideal-gas equation:

$P V = n R T$

to calculate the number of moles of the gas.

We're trying to find the number of moles, $n$, so let's rearrange the equation to solve for $n$:

n = (PV)/(RT)

We know

• P = 755cancel("mm Hg")((1color(white)(l)"atm")/(760cancel("mm Hg"))) = 0.993 $\text{atm}$

• V = 498cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = 0.498 $\text{L}$

• $R = 0.082057 \left(\text{L"·"atm")/("mol"·"K}\right)$

• $T = {100}^{\text{o""C}} + 273 = 373$ $\text{K}$

Plugging these into the equation, we have

n = (PV)/(RT) = ((0.993cancel("atm"))(0.498cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(373cancel("K"))) = color(red)(0.0162 color(red)("mol"

We know that the mass of the vapor is $2.31$ $\text{g}$, so to find the molar mass, we divide this mass by the number of moles:

$\text{molar mass" = "mass"/"moles" = (2.31color(white)(l)"g")/(color(red)(0.0162)color(white)(l)color(red)("mol")) = color(blue)(143"g"/"mol}$

This is the mass of one mole of the vapor. The question asks for its molecular weight, which is the same as its atomic mass. The mass of one mole of a substance is the mass in $\text{amu}$ of one unit of that substance.

Therefore,

overbrace(color(blue)(143"g"/"mol"))^"molar mass" = overbrace(color(blue)(143)color(white)(l)color(blue)("amu"))^"molecular weight"

The molecular weight of the gas is thus color(blue)(143 sfcolor(blue)("[atomic mass](https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance) units".