Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

Expert verified
The mass of argon gas in a \(75.0 \mathrm{mL}\) volume at STP is approximately \(0.134 \mathrm{g}\).

Step by step solution

01

Convert the volume to liters

To start, convert the volume given in milliliters (mL) to liters (L) because STP conditions are usually expressed in liters. This is done by dividing the volume by 1000 as follows: \(75.0 \mathrm{mL} = 75.0/1000 = 0.075 \mathrm{L}\) .
02

Calculate the number of moles of the gas

Next, use the volume and the molar volume at STP to calculate the number of moles of the gas. The molar volume at STP is typically given as \(22.4 \mathrm{L/mol}\) . Therefore the number of moles can be calculated using the formula \( n = V/V_m = 0.075 \mathrm{L} /22.4 \mathrm{L/mol} = 0.00335 \mathrm{mol}\).
03

Calculate the mass of the gas

Finally, use the number of moles and the molar mass of argon to calculate the mass of the gas. The molar mass of argon is around \(40 g/mol\) . The formula is \( m = n \cdot M = 0.00335 \mathrm{mol} \cdot 40 \mathrm{g/mol} = 0.134 \mathrm{g}\).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Conversion
When dealing with gas calculations, understanding how to convert between different volume units is essential, especially from milliliters (mL) to liters (L). This is because gases under Standard Temperature and Pressure (STP) conditions are often measured in liters. Volume conversion is straightforward; you simply divide the volume in milliliters by 1000 to get the volume in liters.
For example, if you have a volume of 75.0 mL, converting it to liters involves the simple calculation:
  • Divide by 1000:
    \(75.0 \text{ mL} = \frac{75.0}{1000} = 0.075 \text{ L}\).
This step ensures that such volumes align with the commonly used literature and scientific calculations when working with gases at STP.
Molar Volume
The concept of molar volume is fundamental when working with gases. At Standard Temperature and Pressure (STP), which is defined as 0°C (273.15 K) and 1 atm pressure, one mole of any ideal gas occupies the same volume, which is 22.4 liters.
The molar volume makes it possible to calculate the amount of substance (in moles) from a gas's measured volume. If you know the volume of a gas at STP, the relationship can be defined as:
  • The formula for finding moles from volume:
    \[ n = \frac{V}{V_m} \]
    where \(n\) is the number of moles, \(V\) is the gas volume, and \(V_m\) is the molar volume (22.4 L/mol at STP).
Understanding molar volume helps in converting between the volume of a gas measured at STP and the number of moles, thereby connecting macroscopic quantities with microscopic chemical identities.
Molar Mass of Argon
Argon, a noble gas, is an element with atomic number 18 and commonly found in the Earth's atmosphere. The molar mass of an element refers to the mass of one mole of its atoms. For argon, the molar mass is roughly 40 g/mol. This value is derived from the atomic mass because each mole contains Avogadro's number of atoms.
To find the mass of a specific quantity of argon, such as in a gas sample, you multiply the number of moles by the molar mass:
  • The formula used:
    \[ m = n \times M \]
    where \(m\) is the mass, \(n\) is the number of moles, and \(M\) is the molar mass of argon (40 g/mol).
Knowing the molar mass is crucial for calculating how much argon is in a given set of conditions, thus facilitating various scientific and industrial applications.

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 4.0 L sample of \(\mathrm{O}_{2}\) gas has a pressure of 1.0 atm. A 2.0 L sample of \(\mathrm{N}_{2}\) gas has a pressure of 2.0 atm. If these two samples are mixed and then compressed in a 2.0 L vessel, what is the final pressure of the mixture? Assume that the temperature remains unchanged.

When solid \(\mathrm{KClO}_{3}\) is heated strongly, it decomposes to form solid potassium chloride, \(\mathrm{KCl}\), and \(\mathrm{O}_{2}\) gas. \(\mathrm{A}\) \(0.415 \mathrm{g}\) sample of impure \(\mathrm{KClO}_{3}\) is heated strongly and the \(\mathrm{O}_{2}\) gas produced by the decomposition is collected over water. When the wet \(\mathrm{O}_{2}\) gas is cooled back to \(26^{\circ} \mathrm{C}\), the total volume is \(229 \mathrm{mL}\) and the total pressure is 323 Torr. What is the mass percentage of \(\mathrm{KClO}_{3}\) in the original sample? Assume that none of the impurities produce oxygen on heating. The vapor pressure of water is 25.22 Torr at \(26^{\circ} \mathrm{C}\).

To establish a pressure of 2.00 atm in a 2.24 L cylinder containing \(1.60 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\) at \(0^{\circ} \mathrm{C},\) (a) add \(1.60 \mathrm{g} \mathrm{O}_{2} ;(\mathrm{b})\) add \(0.60 \mathrm{g} \mathrm{He}(\mathrm{g}) ;(\mathrm{c})\) add \(2.00 \mathrm{g} \mathrm{He}(\mathrm{g})\) (d) release \(0.80 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\)

What is the volume, in milliliters, occupied by \(89.2 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) at \(37^{\circ} \mathrm{C}\) and \(737 \mathrm{mmHg} ?\)

The density of phosphorus vapor is \(2.64 \mathrm{g} / \mathrm{L}\) at \(310^{\circ} \mathrm{C}\) and \(775 \mathrm{mm}\) Hg. What is the molecular formula of the phosphorus under these conditions?

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