Question

The incorrect statement for \(14 \mathrm{~g}\) of \(\mathrm{CO}\) is: (a) It occupies \(2.24\) litre at NTP (b) It corresponds to \(0.5 \mathrm{~mol}\) of \(\mathrm{CO}\) (c) It corresponds to same mol of \(\mathrm{CO}\) and \(\mathrm{N}_{2}\) (d) It corresponds to \(3.01 \times 10^{23}\) molecules of \(\mathrm{CO}\)

Short answer

Statement (a) is incorrect because 14 g of CO occupies 11.2 L at NTP, not 2.24 L.

Step-by-step solution

Identify Molar Mass

Calculate the molar mass of carbon monoxide (CO). The atomic masses are approximately 12 g/mol for carbon (C) and 16 g/mol for oxygen (O), resulting in a molar mass for CO of 12 + 16 = 28 g/mol.

Calculate Moles of CO

Determine the number of moles of CO in a 14 g sample. Use the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{14 \text{ g}}{28 \text{ g/mol}} = 0.5 \text{ mol} \] This confirms statement (b) is correct.

Check Volume at NTP

Calculate the volume of the gas at Normal Temperature and Pressure (NTP), where 1 mole of gas occupies 22.4 L. The volume of 0.5 mol of CO is: \[ 0.5 \times 22.4 = 11.2 \text{ L} \] This indicates statement (a) (2.24 L) is incorrect.

Compare Moles of CO and N2

Since both 0.5 mol of CO and 0.5 mol of nitrogen (N2) imply the same number of molecules or atoms in a mole (Avogadro’s number), statement (c) holds true.

Calculate Number of Molecules

Confirm the number of molecules in 0.5 mol of CO, using Avogadro's number (approximately \(6.02 \times 10^{23}\) molecules per mole): \[ 0.5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}\] Thus, statement (d) is correct.

Key concepts

Moles Calculation

Calculating moles is a fundamental concept in chemistry that helps us understand the amount of a substance. It's crucial when dealing with any chemical reaction or composition. Moles are essentially a bridge between the atomic world and the macro world we interact with every day. When you calculate moles, you're counting atoms or molecules in practical amounts.

To determine the number of moles, you use the formula \( \text{moles} = \frac{\text{mass (in g)}}{\text{molar mass (g/mol)}} \). This formula tells us that by dividing the mass of the substance by its molar mass (the mass of one mole of the substance), we get the number of moles.

• In the case of CO, where the molar mass is 28 g/mol, a 14 g sample results in 0.5 mol.
• Understanding this calculation helps in quantifying the components involved in a reaction.

Knowing the moles is essential in predicting how much product will be formed or needed in a reaction, making it a key step in stoichiometry.

Volume at NTP

Understanding the volume of gases at Normal Temperature and Pressure (NTP) is vital in chemistry, especially when dealing with gas reactions. NTP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm (101.3 kPa). At these conditions, one mole of any gas occupies 22.4 liters.

This uniformity allows chemists to predict gas behavior and reactions with simple calculations. For instance, if we know that 1 mole of a gas occupies 22.4 liters, then 0.5 moles would logically occupy half that volume, which is 11.2 liters.

• This demonstrates that the assertion of CO occupying 2.24 L at NTP is incorrect; hence the accurate volume is actually 11.2 L.

Understanding these gas laws helps in working with gaseous substances in a predictable and safe manner. It simplifies the experimental process and helps in scaling reactions accurately based on the gas volumes involved.

Avogadro's Number

Avogadro's Number, \(6.02 \times 10^{23}\), is a vital constant in chemistry. It represents the number of atoms or molecules in one mole of a substance. This concept is central to the mole concept and helps in converting between atoms, molecules, and moles.

With Avogadro's Number, we can determine the exact number of molecules in a given number of moles. For carbon monoxide (CO) in this context:

• 0.5 moles of CO would contain \(0.5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}\) molecules.

This ability to count particles in a tangible way bridges the gap between macroscopic measurements and microscopic particles. It provides a foundational measure for performing conversions and understanding the scale of chemical reactions.