assessment problem solving an argon gas occupies

V   ––– = k n  

V 1   ––– = k n 1  

V 2   ––– = k n 2  

V 1   V 2 ––– = ––– n 1   n 2

V 1 / n 1 = V 2 / n 2

V 1 n 2 = V 2 n 1

I'll use V 1 n 2 = V 2 n 1 (5.00 L) (1.80 mol) = (x) (0.965 mol) x = 9.33 L (to three sig figs)

2.00 g / 4.00 g/mol = 0.500 mol

V 1 / n 1 = V 2 / n 2 2.00 L / 0.500 mol = 2.70 L / x x = 0.675 mol

0.675 mol − 0.500 mol = 0.175 mol (0.175 mol) (4.00 g/mol) = 0.7 grams of He added

(a) The balloon doubles in volume. (b) The volume of the balloon expands by more than two times. (c) The volume of the balloon expands by less than two times. (d) The balloon stays the same size but the pressure increases. (e) None of the above.

We can perform a calculation using Avogadro's Law: V 1 / n 1 = V 2 / n 2 Let's assign V 1 to be 1 L and V 2 will be our unknown. Let us assign 1 mole for the amount of neon gas and assign it to be n 1 . The mass of argon now added is exactly equal to the neon, but argon has a higher gram-atomic weight (molar mass) than neon. Therefore less than 1 mole of Ar will be added. Let us use 1.5 mol for the total moles in the balloon (which will be n 2 ) after the Ar is added. (I picked 1.5 because neon weighs about 20 g/mol and argon weighs about 40 g/mol.) 1 / 1 = x / 1.5 x = 1.5 answer choice (c).

V 1 / n 1 = V 2 / n 2 5.120 L   18.10 L –––––––– = –––––– 8.500 mol   x x = 30.05 mol 30.05 − 8.500 = 21.55 mol (to four sig figs)

V 1   V 2 ––– = –––– n 1   n 2

214 mL   V 2 ––––––––– = –––––––––– 0.00810 mol   0.00684 mol

V 2 = 181 mL (to three sig figs) When I did the actual calculation for this answer, I used 684 and 810 when entering values into the calculator.

Dividing PV 1 = n 1 RT by PV 2 = n 2 RT, we get V 1 /V 2 = n 1 /n 2 V 2 = V 1 n 2 /n 1 V 2 = [(214 mL) (0.00684 mol)] / 0.00810 mol V 2 = 181 mL In case you don't know, PV = nRT is called the Ideal Gas Law. You'll see it a bit later in your Gas Laws unit, if you haven't already.

PV = nRT V/n = RT / P R is, of course, a constant.

k = RT / P where k is some constant.

V 1 / n 1 = k V 2 / n 2 = k

6.13 / 7.51 = 13.5 / n (6.13) (n) = (13.5) (7.51) n = [(13.5) (7.51)] / 6.13 n = 16.54 mol (this is not the final answer)

16.54 − 7.51 = 9.03 mol (this is the number of moles of gas that were added)

(1.050 mol) (31.9988 g/mol) = 33.59874 g

33.59874 − 7.210 = 26.38874 g

26.38874 g / 31.9988 g/mol = 0.824679 mol

V 1 / n 1 = V 2 / n 2 25.47 L / 1.050 mol = V 2 / 0.824679 mol V 2 = 20.00 L

7.210 g / 31.9988 g/mol = 0.225321 mol

1.050 mol − 0.225321 mol = 0.824679 mol

V 1   V 2 ––––  =  –––– n 1   n 2

V 1   V 2 ––––––––––  =  –––––––––– mass 1 / MM   mass 2 / MM

V 1   V 2 ––––  =  –––– mass 1   mass 2

25.47 L   V 2 ––––––––  =  –––––––– 33.59874 g   26.38874 g V 2 = 20.00 L

P 1 V 1   P 2 V 2 –––––  =  ––––– n 1 T 1   n 2 T 2 Note that it is the full version which includes the moles of gas. Usually a shortened version with the moles not present is used. Since grams are involved (which leads to moles), we choose to use the full version.

V 1   V 2 –––  =  ––– n 1   n 2

V 2 n 1 = V 1 n 2 V 2 = (V 1 ) (n 2 / n 1 )

n = mass / mw V 2 = (V 1 ) [(mass 2 / mw) / (mass 1 / mw)]

V 2 = (V 1 ) (mass 2 / mass 1 )

V 2 = (3.0 L) (5.5 g / 2.2 g) V 2 = 7.5 L

One mole of H 2 molecules has 6.022 x 10 23 H 2 molecules. One mole of H atoms has 6.022 x 10 23 H atoms. The number of independent "particles" in each sample is the same. Therefore, the volumes occupied by the two samples are the same. The volume of the H atoms sample is 20 L. By the way, I agree that one mole of H 2 has twice as many atoms as one mole of H atoms. However, the atoms in H 2 are bound up into one mole of molecules, which means that one molecule of H 2 (with two atoms) counts as one independent "particle" when considering gas behavior.

6.13 L   15.5 L ––––––– = –––––– 8.51 mol   x x = 21.5 mol

21.5 mol − 8.51 mol = 13.0 mol (when properly rounded off)

2.00 L   2.50 L –––––––– = –––––– 0.500 mol   x x = [(0.500 mol) (2.50 L)] / 2.00 L x = 0.625 mol

(4.00 g/mol) (0.125 mol) = 0.500 g Notice that I subtracted 0.500 mol from 0.625 mol and used 0.125 mol in the calculation. This is because I want the amount added, not the final ending amount.

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Chapter 6: Problem 166

A sample of argon gas at STP occupies \(15.0 \mathrm{L} .\) What mass of argon is present in the container?

Short answer, step by step solution, recall the molar volume of a gas at stp, calculate the number of moles of argon gas, recall the molar mass of argon gas, calculate the mass of argon gas, key concepts.

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Molar Volume at STP

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Most popular questions from this chapter

On November \(9,1961,\) the Bell \(\mathrm{X}-15\) test aircraft exceeded Mach \(6,\) or six times the speed of sound. A few weeks later it reached an elevation above 300,000 feet, effectively putting it in space. The combustion of ammonia and oxygen provided the fuel for the X-15. a. Write a balanced chemical equation for the combustion of ammonia given that nitrogen ends up as nitrogen dioxide. b. What ratio of partial pressures of ammonia and oxygen is needed for this reaction?

Generating hydrogen from water or methane is energy intensive. A non-natural enzymatic process has been developed that produces 12 moles of hydrogen per mole of glucose by the reaction: \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 12 \mathrm{H}_{2}(g)+6 \mathrm{CO}_{2}(g)\) What volume of hydrogen could be produced from \(256 \mathrm{g}\) of glucose at STP?

In some aquatic ecosystems, nitrate (NO \(_{3}^{-}\) ) is converted to nitrite \(\left(\mathrm{NO}_{2}^{-}\right),\) which then decomposes to nitrogen and water. As an example of this second reaction, consider the decomposition of ammonium nitrite: $$ \mathrm{NH}_{4} \mathrm{NO}_{2}(a q) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ What would be the change in pressure in a sealed \(10.0 \mathrm{L}\) vessel due to the formation of \(\mathrm{N}_{2}\) gas when the ammonium nitrite in \(1.00 \mathrm{L}\) of \(1.0 \mathrm{MNH}_{4} \mathrm{NO}_{2}\) decomposes at \(25^{\circ} \mathrm{C} ?\)

What happens to the pressure of a gas under the following conditions? a. The absolute temperature is halved and the volume doubles. b. Both the absolute temperature and the volume double. c. The absolute temperature increases by \(75 \%,\) and the volume decreases by \(50 \%\).

Which of the following are not characteristics of an ideal gas? a. The molecules of gas have little volume compared with the volume that they occupy. b. Its volume is independent of temperature. c. The density of all ideal gases is the same. d. Gas atoms or molecules do not interact with one another.

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Question: How would I solve this problem? Argon gas at STP occupies 50.2 liters. Determine the number of moles of argon and the mass in the sample.

How would I solve this problem? Argon gas at STP occupies 50.2 liters. Determine the number of moles of argon and the mass in the sample.

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Given the volume of argon gas and its conditions at STP, use the ideal gas law equation, , to find the number of moles ( ).

Volume of argon = 50.2 L At STP, temperature = 273.15 K  …

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