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What is the pressure exerted by 2.10 g Ar gas at 18.0 °C in a 660-ml flask?

1.91*10^-3 atm
4.71 atm
1.91 atm
0.118 atm
76.1 atm

User Dukasvili
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1 Answer

3 votes

Answer:

1.91 atm

Step-by-step explanation:

To start this question I would first look at what the question is giving us:

M= 2.10g Argon

T= 18.0°C

V= 660-mL

P= ?atm

We now would recognize that from the things given, the formula that suits it would be PV=nRT because it would have the given data present were we can solve for P.

R is a constant that is 0.082 L*atm/K*mol. This constant lets us know what units we need to convert in order move forward.

We would need our volume to be in Liters, our pressure in atm, our temperature in K and our mass in mols.

So then in order to convert the mass given, we would need to go from grams to mols. We do this by starting off with our given mass (2.10g) and then dividing it by the molar mass of the element Argon (39.948g). This would make a total of 0.052mols.

To convert the Celsius to Kelvin, we just need to add 273 to the temperature given. 18.0°C + 273 = 291K.

To convert our volume, 1000mL=1L, so we would divide our given mL by 1000. 600/1000= 0.660L.

Now we can go ahead and use our formula and plug in the numbers to solve for P.

PV=nRT

(x atm)(0.660 L)=(0.0526 mol)(0.082 L*atm/K*mol)(291 K)

When solved, we would get a final answer of 1.90173 atm. Our closest answer being 1.91 atm.

User Raju Ugale
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6.7k points