Question

Hello, Can someone please help me solve this problem? An experiment shows that a 114 mL gas sample has a mass of 0.161 g at a pressure of 713 mmHg and a temperature of 26 ∘C. *WHAT IS THE MOLAR MASS OF THE GAS?*

use sig figs

Answer 1

Alright I will use sig figs...see below
`37.0 g/(mol)`

Explanation:

Ask yourself: What are the units of molar mass?
`g/(mol)` correct?
We will do dimensional analysis in a way we end up with `g/(mol)` as our final unit, we will utilize the gas law constant: `0.08205 (L*atm)/(mol*K)`, so let us first convert into a constant with mmHg instead of atm:

So:
#(0.08205 L*cancel(atm))/(mol*K)*(760 mmHg)/ (1 cancel(atm))= (62.4L*mmHg)/(mol*K) #

So:
`0.161 g* (62.4cancel(L)*cancel(mmHg))/(mol*cancel(K))*(299 cancel(K))/ (0.114cancel(L))*1/(713 cancel(mmHg))= 37.0 g/(mol)`

Answer 2

Here, we want to relate the moles present to the molar mass given its mass.

Recall,

`PV = nRT`

Consider that,

`=> n = (PV)/(RT)`

`n=(713"mmHg" * (1"atm")/(760"mmHg") * 114"mL" * "L"/(10^3"mL"))/((0.08206"L" * "atm")/("mol" * "K") * 299.15"K") approx 4.357*10^-3"mol"`

of this unknown gas is present.

Hence,

`"MM" = m/n = (0.161"g")/(4.357*10^-3"mol") approx (36.9_5"g")/"mol"`

is the approximate molar mass of this gas, and we get `ul"37.0 g/mol"` to three sig figs.