Question e9ae8

Dec 11, 2016

(77.95" g")/"mole"

Explanation:

Let $R =$ The Gas Constant $= 62.363$ L Torr K"^-1 mole"^-1

Given: $P = 400 \text{ torr}$ and $T = 127 \text{ C"^@ = 400" } K$

The Ideal Gas law is:

$P V = n R T$

We are not given the volume but we do not need it; the reason will soon become obvious. Solve for the Volume per Moles:

$\frac{V}{n} = \frac{R T}{P}$

V/n = ((62.363" L Torr K"^-1" mole"^-1)(400" K"))/(400" Torr")

I am taking an extra step to show you that the Numbers for temperature and pressure cancel and and their units are canceled by the gas constant:

$\frac{V}{n} = \left(\left(62.363 \text{ L "cancel("Torr")cancel(K^-1)" mole"^-1)(cancel(400)cancel(" K")))/(cancel(400)cancel(" Torr}\right)\right)$

No matter what Volume or number of moles we pick, at this temperature and pressure, the following ratio is true:

(62.363" L")/"mole" = V/n

We are given the mass density: (1.25" g")/"L"

To find the Molar Mass, we use the Factor-Label Method (a.k.a. Dimensional Analysis):

(1.25" g")/cancel("L")(62.363cancel(" L"))/"mole" = 77.95" g"/"mole"#