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Chapter 5: Problem 36

You have \(357 \mathrm{~mL}\) of chlorine trifluoride gas at \(699 \mathrm{mmHg}\) and \(45^{\circ} \mathrm{C}\). What is the mass (in \(\mathrm{g}\) ) of the sample?

Short Answer

Expert verified
1.17 g

Step by step solution

01

Convert the given values to standard units

First, convert the volume from \(\text{mL}\) to \(\text{L}\) and the temperature from \(^\text{C}\) to \(K\). \[ 357 \text{ mL} = 0.357 \text{ L} \] and \[ 45^\text{C} = 45 + 273.15 = 318.15 \text{ K}\]
02

Identify and write down the formula

The ideal gas law equation is \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature.
03

Convert pressure to atmospheres

Convert \(699 \text{ mmHg}\) to atmospheres using the conversion factor \(1 \text{ atm} = 760 \text{ mmHg}\). \[ 699 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 0.92 \text{ atm}\]
04

Solve for moles of gas

Use the ideal gas law to solve for \(n\). \[n = \frac{PV}{RT} = \frac{(0.92 \text{ atm})(0.357 \text{ L})}{(0.0821 \frac{\text{L} \text{ atm}}{\text{mol} \text{ K}})(318.15 \text{ K})} = 0.01269 \text{ moles} \]
05

Convert moles to grams

Find the molar mass of \( \text{ClF}_3 \). The molar mass of \( \text{ClF}_3 \) is \( (1 \times 35.45) + (3 \times 19.00) = 92.45 \text{ g/mol}\). Calculate the mass: \[ 0.01269 \text{ moles} \times 92.45 \frac{\text{g}}{\text{mol}} = 1.17 \text{ g} \]

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Key Concepts

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

gas laws
Gas laws are mathematical relationships that describe the behavior of gases. One of the most important gas laws is the Ideal Gas Law, represented by the equation:

\(PV = nRT\).

Here, \(P\) is the pressure of the gas, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature.

The Ideal Gas Law combines several simpler gas laws:
  • Boyle's Law: Describes the inverse relationship between pressure and volume, keeping temperature constant \( (PV = k)\).
  • Charles's Law: Describes the direct relationship between volume and temperature, keeping pressure constant \( (V \, \text{is directly proportional to} \, T)\).
  • Avogadro's Law: States that equal volumes of gases at the same temperature and pressure contain the same number of moles \(V \, \text{is directly proportional to } \, n)\).
Understanding these relationships helps to predict how a gas will behave under different conditions.
mole calculation
Calculating the number of moles in a gas involves using the formula \(n = \frac{PV}{RT}\), derived from the Ideal Gas Law.

In the given exercise, we have:
  • Pressure \( (P) = 0.92\, \text{atm}\)
  • Volume \( (V) = 0.357\, L\)
  • Ideal Gas Constant \( (R) = 0.0821\, \frac{L \, atm}{mol \, K} \)
  • Temperature \( (T) = 318.15\, K \)
Plugging these values into the formula, we get:

\[ n = \frac{(0.92 \, atm)(0.357 \, L)}{(0.0821 \, \frac{L \, atm}{mol \, K})(318.15 \, K)} = 0.01269 \, \text{moles} \]

This tells us how many moles of chlorine trifluoride gas are present in the given conditions.
unit conversion
Unit conversion is crucial to ensure that all measurements are in compatible units when applying mathematical formulas.

In this exercise, we needed to convert:
  • Volume from \(357\,mL\) to \(0.357\,L\). Remember, \(1\,L = 1000\,mL\).
  • Temperature from \(45^{\circ}C\) to \(318.15\,K\). Use the formula \(T(K) = T(^\circ C) + 273.15\).
  • Pressure from \(699\,mmHg\) to \(0.92\,atm\). Here, \(1\,atm = 760\,mmHg\), so the conversion factor is \(\frac{1 atm}{760 mmHg}\).
These conversions align all units into the standard form required by the Ideal Gas Law.
molar mass
Molar mass is the mass of one mole of a substance. It is usually expressed in grams per mole \( g/mol \).

For chlorine trifluoride \(ClF_3\), we need to calculate it based on the atomic masses of the constituent elements:
  • Chlorine (Cl): \(35.45\,g/mol\)
  • Fluorine (F): \(19.00\,g/mol\)
Chlorine trifluoride has 1 chlorine atom and 3 fluorine atoms, so:

\[Molar\, mass = (1 \times 35.45) + (3 \times 19.00) = 92.45 \, g/mol\]

Finally, we convert moles to grams using this molar mass. Given \(0.01269\, \text{moles}\), the mass in grams is:

\[Mass = 0.01269 \,mol \times 92.45\, \frac{g}{mol} = 1.17\,g\] This result shows the mass of the chlorine trifluoride gas sample.

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Most popular questions from this chapter

Three \(5-\mathrm{L}\) flasks, fixed with pressure gauges and small valves, each contain \(4 \mathrm{~g}\) of gas at \(273 \mathrm{~K}\). Flask A contains \(\mathrm{H}_{2}\), flask B contains He, and flask C contains \(\mathrm{CH}_{4}\). Rank the flask contents in terms of (a) pressure, (b) average kinetic energy of the particles, (c) diffusion rate after the valve is opened, (d) total kinetic energy of the particles, (e) density, and (f) collision frequency.

What is the effect of the following on the volume of \(1 \mathrm{~mol}\) of an ideal gas? (a) The pressure is tripled (at constant \(T\) ). (b) The absolute temperature is increased by a factor of 3.0 (at constant \(P\) ). (c) Three more moles of the gas is added (at constant \(P\) and \(T\) ).

Will the volume of a gas increase, decrease, or remain unchanged with each of the following sets of changes? (a) The pressure is decreased from 2 atm to 1 atm, while the temperature is decreased from \(200^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C} .\) (b) The pressure is increased from 1 atm to 3 atm, while the temperature is increased from \(100^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\). (c) The pressure is increased from 3 atm to 6 atm, while the temperature is increased from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). (d) The pressure is increased from 0.2 atm to 0.4 atm, while the temperature is decreased from \(300^{\circ} \mathrm{C}\) to \(150^{\circ} \mathrm{C}\).

Convert the following: (a) 0.745 atm to \(\mathrm{mmHg}\) (b) 992 torr to bar (c) \(365 \mathrm{kPa}\) to atm (d) \(804 \mathrm{mmHg}\) to \(\mathrm{kPa}\)

A person inhales air richer in \(\mathrm{O}_{2}\) and exhales air richer in \(\mathrm{CO}_{2}\) and water vapor. During each hour of sleep, a person exhales a total of about \(300 \mathrm{~L}\) of this \(\mathrm{CO}_{2}\) -enriched and \(\mathrm{H}_{2} \mathrm{O}\) -enriched air. (a) If the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) in exhaled air are each 30.0 torr at \(37.0^{\circ} \mathrm{C},\) calculate the mass \((\mathrm{g})\) of \(\mathrm{CO}_{2}\) and of \(\mathrm{H}_{2} \mathrm{O}\) exhaled in \(1 \mathrm{~h}\) of sleep. (b) How many grams of body mass does the person lose in \(8 \mathrm{~h}\) of sleep if all the \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) exhaled come from the metabolism of glucose? $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$
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