Problem 65
Question
The number of moles of oxygen in one litre of air containing \(21 \%\) oxygen by volume, in standard conditions, is (a) \(0.176 \mathrm{~mol}\) (b) \(0.32 \mathrm{~mol}\) (c) \(0.0093 \mathrm{~mol}\) (d) \(2.20 \mathrm{~mol}\)
Step-by-Step Solution
Verified Answer
The correct answer is (c) 0.0093 mol.
1Step 1: Determine the Volume of Oxygen
Since the air contains 21% oxygen by volume, in one litre (1000 mL) of air, the volume of oxygen present is 0.21 liters. This is found by multiplying the total volume (1 L) by the percentage of oxygen (0.21).
2Step 2: Use Molar Volume to Find Moles
Under standard conditions (STP), one mole of gas occupies a volume of 22.4 liters. Therefore, the number of moles of oxygen can be calculated by dividing the volume of oxygen by the molar volume: \[\text{Moles of } O_2 = \frac{0.21 \text{ L}}{22.4 \text{ L/mol}}.\]
3Step 3: Calculate the Number of Moles
Perform the calculation: \[\text{Moles of } O_2 = \frac{0.21}{22.4} \approx 0.009375 \text{ mol}.\] This result can be approximated to three significant figures as 0.0093 mol.
Key Concepts
Gas LawsStandard Temperature and PressureVolume to Moles Conversion
Gas Laws
Gas laws are essential to understanding how gases behave under different conditions of temperature and pressure. One of the most well-known laws is the ideal gas law formula, which is states as \(PV = nRT\). In this formula, \(P\) represents pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. This equation is a powerful tool to relate the four fundamental properties of gases under ideal conditions.
Using gas laws, we can predict how a gas will respond to changes in temperature or pressure, which aligns perfectly with our task of determining the moles of gas under standard conditions. Understanding these laws goes beyond simply memorizing formulas. It's about recognizing that gas molecules are constantly moving and colliding with each other and their container, which directly impacts the pressure and volume observed.
Using gas laws, we can predict how a gas will respond to changes in temperature or pressure, which aligns perfectly with our task of determining the moles of gas under standard conditions. Understanding these laws goes beyond simply memorizing formulas. It's about recognizing that gas molecules are constantly moving and colliding with each other and their container, which directly impacts the pressure and volume observed.
Standard Temperature and Pressure
Standard Temperature and Pressure (STP) is a set of conditions that allows scientists to work with gases in a consistent way. At STP, temperature is set at 0°C or 273.15 Kelvin, and pressure is standardized at 1 atmosphere (atm). These conditions offer a base for comparing gas volumes and behaviors.
One critical aspect of STP is that, under these conditions, one mole of an ideal gas occupies 22.4 liters. This is termed as the molar volume of a gas at STP. It's handy for converting volume to moles, as seen in our exercise where knowing that 1 mole equals 22.4 liters at STP allowed us to find the number of moles of oxygen efficiently. By setting these conditions, calculations become standardized, ensuring that scientists across different laboratories can replicate and verify results easily.
One critical aspect of STP is that, under these conditions, one mole of an ideal gas occupies 22.4 liters. This is termed as the molar volume of a gas at STP. It's handy for converting volume to moles, as seen in our exercise where knowing that 1 mole equals 22.4 liters at STP allowed us to find the number of moles of oxygen efficiently. By setting these conditions, calculations become standardized, ensuring that scientists across different laboratories can replicate and verify results easily.
Volume to Moles Conversion
Converting from volume to moles is a straightforward process thanks to the molar volume concept. Molar volume is the volume occupied by one mole of a gas at a given temperature and pressure, usually 22.4 liters at STP. This conversion is crucial when determining amounts of substances in chemical reactions.
In our exercise, we calculated the volume of oxygen in the air first, discovering it was 0.21 liters. Using the molar volume, we simply divided this by 22.4 liters per mole to find the number of moles of oxygen. The resulting calculation was \(\frac{0.21 \text{ L}}{22.4 \text{ L/mol}}\), equaling approximately 0.009375 moles, which simplifies understanding the quantities involved in chemical processes. This process underscores the importance of understanding the relation between volume and moles, reduction complexity and increasing accuracy in chemistry computations.
In our exercise, we calculated the volume of oxygen in the air first, discovering it was 0.21 liters. Using the molar volume, we simply divided this by 22.4 liters per mole to find the number of moles of oxygen. The resulting calculation was \(\frac{0.21 \text{ L}}{22.4 \text{ L/mol}}\), equaling approximately 0.009375 moles, which simplifies understanding the quantities involved in chemical processes. This process underscores the importance of understanding the relation between volume and moles, reduction complexity and increasing accuracy in chemistry computations.
Other exercises in this chapter
Problem 62
If \(\mathrm{N}_{\mathrm{A}}\) is Avogadro's number then number of valence electrons in \(4.2 \mathrm{~g}\) of nitride ions \(\left(\mathrm{N}^{3}\right)\) is (
View solution Problem 63
In the final answer of the expression $$ \frac{(29.2-20.2)\left(1.79 \times 10^{5}\right)}{1.37} $$ The number of significant figures is (a) 2 (b) 4 (c) 6 (d) 7
View solution Problem 66
Number of atoms in \(4.25 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) is approximately (a) \(6 \times 10^{23}\) (b) \(15 \times 10^{23}\) (c) \(1.5 \times 10^{23}\) (d
View solution Problem 68
The percentage weight of \(\mathrm{Zn}\) in white vitriol \(\left[\mathrm{ZnSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\right]\) is approximately equal to \((\mat
View solution