\(\mathrm{Kr}(\mathrm{g})\) in a 18.5 L cylinder exerts a pressure of \(11.2 \mathrm{atm}\) at \(28.2^{\circ} \mathrm{C} .\) How many grams of gas are present?

Short Answer

Expert verified
The mass of the Krypton gas in the cylinder is 73.24 grams.

Step by step solution

01

Identify Given Values

The values given in the problem are Pressure(P) = 11.2 atm, Volume(V) = 18.5 L, Temperature(T) = \(28.2^{\circ} \mathrm{C}\) (which needs to be converted to Kelvin by adding 273.15), and the gas constant(R) = 0.0821 L.atm/K.mol.
02

Convert Temperature to Kelvin

The temperature is given in degrees Celsius but the ideal gas law requires temperature to be in Kelvin. Therefore, convert the temperature to Kelvin by adding 273.15 to the Celsius temperature. This gives \(T = 28.2 + 273.15 = 301.35 \mathrm{K}\).
03

Substitute values into the Ideal gas law

Substitute the values of the pressure, volume, gas constant, and temperature into the ideal gas law (\(PV = nRT\)). This gives \(11.2 \mathrm{atm} * 18.5 \mathrm{L} = n * 0.0821 * 301.35 \mathrm{K}\). Solve for n.
04

Solve for n

Solve the equation for n, the number of moles. \( n = \frac{11.2 \mathrm{atm} * 18.5 \mathrm{L}}{0.0821 * 301.35 \mathrm{K}} = 0.873 \mathrm{moles}\).
05

Convert moles to grams

Convert the number of moles to grams using the molar mass of Krypton, which is 83.8 g/mol. This gives \(\mathrm{Mass} = 0.873 \mathrm{moles} * 83.8 \mathrm{g/mol} = 73.24 \mathrm{g}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 0.168 L sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(26^{\circ} \mathrm{C}\) and a barometric pressure of \(737 \mathrm{mm} \mathrm{Hg}\). In the gas that is collected, what is the percent water vapor (a) by volume; (b) by number of molecules; (c) by mass? (Vapor pressure of water at \(26^{\circ} \mathrm{C}=25.2 \mathrm{mmHg}\).)

A sample of \(\mathrm{N}_{2}(\mathrm{g})\) effuses through a tiny hole in \(38 \mathrm{s}\) What must be the molar mass of a gas that requires \(64 \mathrm{s}\) to effuse under identical conditions?

The amount of ozone, \(\mathrm{O}_{3}\), in a mixture of gases can be determined by passing the mixture through a solution of excess potassium iodide, KI. Ozone reacts with the iodide ion as follows: $$\begin{aligned} \mathrm{O}_{3}(\mathrm{g})+3 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) & \longrightarrow \\ \mathrm{O}_{2}(\mathrm{g})+\mathrm{I}_{3}^{-}(\mathrm{aq}) &+2 \mathrm{OH}^{-}(\mathrm{aq}) \end{aligned}$$ The amount of \(I_{3}^{-}\) produced is determined by titrating with thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}:\) $$\mathrm{I}_{3}^{-}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \longrightarrow 3 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})$$ A mixture of gases occupies a volume of \(53.2 \mathrm{L}\) at \(18^{\circ} \mathrm{C}\) and \(0.993 \mathrm{atm} .\) The mixture is passed slowly through a solution containing an excess of KI to ensure that all the ozone reacts. The resulting solution requires \(26.2 \mathrm{mL}\) of \(0.1359 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) to titrate to the end point. Calculate the mole fraction of ozone in the original mixture.

Consider the statements (a) to (e) below. Assume that \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) behave ideally. State whether each of the following statements is true or false. For each false statement, explain how you would change it to make it a true statement. (a) Under the same conditions of temperature and pressure, the average kinetic energy of \(\mathrm{O}_{2}\) molecules is less than that of \(\mathrm{H}_{2}\) molecules. (b) Under the same conditions of temperature and pressure, \(\mathrm{H}_{2}\) molecules move faster, on average, than \(\mathrm{O}_{2}\) molecules. (c) The volume of \(1.00 \mathrm{mol}\) of \(\mathrm{H}_{2}(\mathrm{g})\) at \(25.0^{\circ} \mathrm{C}\) 1.00 atm is \(22.4 \mathrm{L}\) (d) The volume of \(2.0 \mathrm{g} \mathrm{H}_{2}(\mathrm{g})\) is equal to the volume of \(32.0 \mathrm{g} \mathrm{O}_{2}(\mathrm{g}),\) at the same temperature and pressure. (e) In a mixture of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) gases, with partial pressures \(P_{\mathrm{H}_{2}}\) and \(P_{\mathrm{O}_{2}^{\prime}}\) respectively, the total pressure is the larger of \(P_{\mathrm{H}_{2}}\) and \(P_{\mathrm{O}_{2}}\).

What is the mass of argon gas in a \(75.0 \mathrm{mL}\) volume at STP?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free