Question

At \(0^{\circ} \mathrm{C}\) and 0.500 atm, 4.48 L of gaseous \(\mathrm{NH}_{3}\) (a) contains \(6.02 \times 10^{22}\) molecules; (b) has a mass of \(17.0 \mathrm{g} ;(\mathrm{c})\) contains \(0.200 \mathrm{mol} \mathrm{NH}_{3} ;\) (d) has a mass of \(3.40 \mathrm{g}\).

Short answer

The answers are: (a) 0.1 mol, (b) 0.998 mol, (c) 0.200 mol, (d) 0.200 mol.

Step-by-step solution

Calculations for part (a)

We know that at \(0^{\circ} \mathrm{C}\) and 0.500 atm, 4.48 L of gaseous NH_3 contains \(6.02 \times 10^{22}\) molecules. Since \(6.02 \times 10^{23}\) molecules form a mole, thus \(6.02 \times 10^{22}\) molecules will form \(6.02 \times 10^{22} / 6.02 \times 10^{23} = 0.1\) mole of NH_3.

Calculations for part (b)

The molar mass of NH_3 is 14.007 (Nitrogen) + 3*1.008 (Hydrogen) = 17.031 g/mol. So, 17.0 g of NH_3 is equal to \(17.0 g / 17.031 g/mol = 0.998\) moles.

Answer for part (c)

As no calculations are required here, we just state that under the conditions given, 4.48 L of gaseous NH_3 contains 0.200 mol.

Calculations for part (d)

Since the molar mass of NH_3 is 17.031 g/mol, thus 3.40 g of NH_3 is equal to \(3.40 g / 17.031 g/mol = 0.200\) moles.

Key concepts

Molar Mass

The concept of molar mass is crucial in understanding the relationships between grams, moles, and molecules of a substance. Molar mass is defined as the mass of one mole of a substance and it is measured in grams per mole (g/mol).
To find the molar mass of a compound like ammonia (H_3), simply add the atomic masses of all the atoms present in the molecule. For NH_3:

• Nitrogen (N) has an atomic mass of approximately 14.007 g/mol.
• Hydrogen (H) has an atomic mass of about 1.008 g/mol.

The calculation becomes: 14.007 + 3(1.008) = 17.031 g/mol.Understanding the molar mass allows you to convert between grams and moles. For instance, 17.0 g of NH_3 is \(17.0 \text{ g} / 17.031 \text{ g/mol} \approx 0.998 \text{ moles}\).This calculation converts mass to moles, a fundamental step in many chemical equations and reactions.

Moles and Molecules

Understanding moles and molecules is key to grasping the quantitative aspects of chemistry. A "mole" is a unit in chemistry that measures a specific number of entities, usually atoms or molecules, similar to how a "dozen" represents 12 items.
In chemistry, one mole contains Avogadro's number of entities, which is approximately \(6.02 \times 10^{23}\).

• This means that 1 mole of N_3 contains \(6.02 \times 10^{23}\) molecules of N_3.

When you know the number of molecules, you can determine the number of moles by dividing the given molecule count by Avogadro's number.For example, \(6.02 \times 10^{22}\) molecules would equal \(6.02 \times 10^{22} / 6.02 \times 10^{23} = 0.1 \text{ mole}\).This concept allows chemists to quantify substances in a consistent and standard way, facilitating easier calculations and comparisons between different chemical substances.

Volume and Pressure Relationships

Volume and pressure relationships are important aspects of gases, described by the Ideal Gas Law, which is \(PV = nRT\).Here:

• \(P\) stands for Pressure, typically in atmospheres (atm).
• \(V\) is the Volume, often measured in liters (L).
• \(n\) represents the number of moles of the gas.
• \(R\) is the Ideal Gas Constant, which is \(0.0821 \text{ L atm/mol K}\), a standard value.
• \(T\) is the Temperature in Kelvin (K).

At standard conditions, it's useful to understand how changing pressure impacts the volume and quantity represented by mole concept.In the example provided, at a fixed temperature of \(0^{\circ}C\) and pressure of 0.500 atm, 4.48 L of N_3 are considered. Here, no direct use of the Ideal Gas Law was needed for certain calculations, yet it illustrates how crucial understanding these relationships are in practical scenarios.In reality, altering any of these simple variables—such as reducing the volume while maintaining temperature—solutions and calculations can determine other values like moles or total molecules, crucial for laboratory and industrial applications.