/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Determine the molecular weight o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the molecular weight of a gas if \(4.50 \mathrm{~g}\) of it occupies \(4.0\) liters at 950 torr \(\mathrm{Hg}\) and \(182^{\circ} \mathrm{C}\). \(\mathrm{r}=.082\) liter \(-\mathrm{atm} / \mathrm{mole}-{ }^{\circ} \mathrm{K}\)

Short Answer

Expert verified
The molecular weight of the gas is approximately 27.11 g/mol.

Step by step solution

01

Convert the given temperature to Kelvin

To convert the given temperature from Celsius to Kelvin, we can use the following equation: \(T(K) = T(°C) + 273.15\) Where: - \(T(K)\) is the temperature in Kelvin - \(T(°C)\) is the temperature in Celsius Given the temperature in Celsius is 182°C, we can calculate the temperature in Kelvin: \( T(K) = 182 + 273.15 = 455.15 \mathrm{~K} \)
02

Convert the given pressure to atmospheres

The pressure is given in torr, but we need it in atmospheres for our calculation. To convert the pressure to atmospheres, we can use the following equation: \(P(atm) = P(torr) \times \frac{1 \mathrm{~atm}}{760 \mathrm{~torr}}\) Where: - \(P(atm)\) is the pressure in atmospheres - \(P(torr)\) is the pressure in torr Given the pressure is 950 torr, we can calculate the pressure in atmospheres: \( P(atm) = 950 \times \frac{1}{760} = 1.25 \mathrm{~atm} \)
03

Find the number of moles of the gas using the Ideal Gas Law

The Ideal Gas Law equation is given as follows: \(PV = nRT\) Where: - \(P\) is the pressure in atmospheres - \(V\) is the volume in liters - \(n\) is the number of moles of the gas - \(R\) is the gas constant in \(L\mathrm{~atm}/\mathrm{mol~K}\) - \(T\) is the temperature in Kelvin We are given the values of \(P\), \(V\), \(R\), and \(T\), so we can solve for \(n\): \(n = \frac{PV}{RT}\) Using the values obtained in Steps 1 and 2: \( n = \frac{1.25 \times 4}{0.082 \times 455.15} \) Calculating the number of moles: \( n = 0.166 \mathrm{~mol} \)
04

Calculate the molecular weight of the gas

The molecular weight of the gas can be calculated using the following equation: Molecular weight (MW) = \(\frac{Mass}{n}\) Where: - Mass is the given mass of the gas - \(n\) is the number of moles Using the given mass of 4.50 g and the number of moles calculated in Step 3: MW = \(\frac{4.50}{0.166}\) Calculating the molecular weight: MW ≈ 27.11 g/mol The molecular weight of the gas is approximately 27.11 g/mol.

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 91Ó°ÊÓ!

Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry and physics, which relates the pressure, volume, temperature, and amount of an ideal gas. Represented by the equation \( PV = nRT \), it allows us to solve for any one of the variables when the others are known.

For instance, in our textbook exercise, the 'number of moles' (n) of a gas is unknown. The Ideal Gas Law rearranges to \( n = \frac{PV}{RT} \) to find this quantity. While working with this law, it's crucial to ensure that units are consistent – pressures in atmospheres, volume in liters, temperature in Kelvin, and the gas constant (R) with appropriate units (usually \( L\mathrm{~atm}/\mathrm{mol~K} \)).

This law proves especially useful when dealing with gaseous compounds in chemical reactions, determining molar mass, and even in more advanced applications such as calculating gas flows and understanding thermodynamic processes.
Convert Celsius to Kelvin
To work accurately with the Ideal Gas Law, temperature must be measured on an absolute scale – the Kelvin scale. Converting Celsius to Kelvin is a simple yet critical step in many scientific calculations. The formula \( T(K) = T(°C) + 273.15 \) is used to make this conversion, ensuring temperature values do not result in negative numbers which have no physical meaning in terms of the kinetic energy of particles.

Thinking of Kelvin as Celsius shifted upward by 273.15 units provides a mental shortcut for this conversion. For our exercise, the temperature given was \(182°C\), which we converted to \(455.15 K\). This temperature in Kelvin signifies the absolute temperature required for the equation to hold true, as the Ideal Gas Law is temperature-dependent.
Convert torr to atmospheres
Pressure measurements come in various units depending on the context, so conversions are often necessary. In the context of ideal gases, we commonly use atmospheres (atm), but lab measurements might be in torr or millimeters of mercury (mmHg). The conversion factor between torr and atmospheres is a constant: \(1 atm = 760 torr\).

To convert from torr to atm, you divide the torr value by 760, illustrated in our textbook exercise with \( P(atm) = P(torr) \times \frac{1}{760} \). Such conversions ensure compatibility with the Ideal Gas Law's parameters, since the gas constant (R) is typically defined in terms of atmospheres.
Moles calculation
The mole is a fundamental unit in chemistry that quantifies the amount of a substance. Moles provide a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. In ideal gas calculations like in our exercise, once temperature and pressure are converted to their appropriate units, we determine the number of moles (n) of gas present using the rearranged Ideal Gas Law.In the exercise scenario, after determining the pressure in atmospheres and temperature in Kelvin, we computed the mole quantity by \( n = \frac{PV}{RT} \). Substituting the appropriate values, we could solve for n. After finding the moles, we used them to calculate the molecular weight, essentially allowing us to understand the gas's chemical makeup based on how much of it fills up a certain space under known conditions.

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

Of the following two pairs, which member will more likely deviate from ideal gas behavior? (1) \(\mathrm{N}_{2}\) versus \(\mathrm{CO},(2)\) \(\mathrm{CH}_{4}=\) versus \(\mathrm{C}_{2} \mathrm{H}_{6}\).

Assuming ideal gas behavior, what is the molecular weight of a gas with a density of \(2.50 \mathrm{~g} /\) liter at \(98^{\circ} \mathrm{C}\) and \(.974 \mathrm{~atm}\) ?

A research worker isolated an unknown compound from one of his reaction products. This compound was a volatile liquid with a boiling point of \(65^{\circ} \mathrm{C}\). A.134 \(\mathrm{g}\) sample of the liquid was vaporized and the gas was collected. The temperature of the collecting chamber was kept at \(98^{\circ} \mathrm{C}\) in order that liquefaction would not occur. After all the sample was vaporized, the volume of the gas was measured as \(.0532\) liters at \(98^{\circ} \mathrm{C}\) and \(737 \mathrm{~mm} .\) Calculate the molecular weight.

On a hot day, the pressure in an automobile tire increases. Assuming that the air in a tire at \(59^{\circ} \mathrm{F}\) increases in pressure from \(28.0 \mathrm{lbs} / \mathrm{in}^{2}\) to \(30.0 \mathrm{lbs} / \mathrm{in}^{2},(\mathrm{a})\) what is the temperature of the air in the tire, assuming no change in volume? (b) what will the pressure be if the temperature rises to \(106^{\circ} \mathrm{F}\) ?

Three researchers studied 1 mole of an ideal gas at \(273^{\circ} \mathrm{K}\) in order to determine the value of the gas constant, \(R\). The first researcher found that at a pressure of 1 atm the gas occupies \(22.4 \ell\). The second researcher found that the gas occupies \(22.4 \ell\) at a pressure of 760 torr. Finally, the third researcher reported the product of pressure and volume as 542 cal. What value for \(\mathrm{R}\) did each researcher determine?

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.