# chemistry SATP Help?????

so on a work sheet I got this question: A sample liquid is placed in a flask and heated to 47oc to vaporize it. The volume of the flask is 100.00 Ml; after weighing it is found to have .250g of vapor atmospheric pressure in the lab is 102 Kpa. using the combined gas law find the volume of gas at SATP, I don't have the second temperature so I am really confused how to set this question up. Please help soon because my quiz is tomorrow thanks :)

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• 9 years ago

I will show 2 methods to solve for the molar mass

.250g is the mass of liquid before it vaporized

STP means standard temperature and pressure. Standard temperature is 0˚C = 273˚K, and standard pressure is 1 atmosphere.

http://en.wikipedia.org/wiki/Pascal_(unit)

The information below is from the website above.

On Earth, standard atmospheric pressure is 101.325 kPa.

Now you have the 2 numbers that you needed.

I will show you how to use these numbers to get the answer.

(Pressure in atmospheres) * (Volume in liters) = (Number of moles) * (gas constant) * (Temperature in ˚K)

Number of moles = mass ÷ molar mass

(Pressure in atmospheres) * (Volume in liters) = (mass/molar mass) * (gas constant) * (Temperature in ˚K)

Pressure in atmospheres = 102/101.325 atm

Volume in liters = 0.1 L

Mass = 0.250 g

Gas constant = 22.4/273 = 0.08205

Temperature = 47 + 273 = 320˚K

(102/101.325) * 0.1 = (0.250/molar mass) * 0.08205 * 320

Molar mass = (0.250 * 0.08205 * 320) ÷ [(102/101.325) * 0.1]

Molar mass = 65.2 grams

OR

(P1 * V1)/ T1 = (101.325 * V stp) / 273

P1 = 102

V1 = 0.1 L

T1 = 47 + 273 = 320˚K

(102 * 0.1)/ 320 = (101.325 * V stp) / 273

V stp = (102 * 0.1 * 273) ÷ (320 * 101.325) = 0.08588 Liter

At STP, 1 mole of a gas has a volume of 22.4 liters.

Number of moles = 0.08588 ÷ 22.4 = 0.003834

Mass of 1 mole = mass ÷ number of moles = 0.250 ÷ 0.003834

Molar mass = 65.2 grams