Question #e9ae8

Question

1905 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-11T17:42:59

$\frac{77.95 g}{mole}$ $(77.95" g")/"mole"$

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

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

$P V = n R T$ $PV = nRT$

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

$\frac{V}{n} = \frac{R T}{P}$ $V/n = (RT)/P$

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

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:

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

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):

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