Problem 42 - A Complete Resource Book in Chemistry for JEE Main

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 \(\mathrm{CO}\)

Step-by-Step Solution

Answer

The incorrect statement is (a) because 0.5 mol of CO occupies 11.2 litres, not 2.24 litres.

Step 1: Calculate moles of CO

The molar mass of CO is calculated as follows: Carbon (C) has a molar mass of \(12 \text{ g/mol}\), and Oxygen (O) has a molar mass of \(16 \text{ g/mol}\). Thus, the molar mass of \(\mathrm{CO}\) is \(12 + 16 = 28 \text{ g/mol}\). For \(14 \text{ g}\) of \(\mathrm{CO}\), the number of moles is \( \frac{14}{28} = 0.5 \text{ mol}\).

Step 2: Verify volume occupied at NTP

At Normal Temperature and Pressure (NTP), 1 mole of a gas occupies \(22.4 \text{ litres}\). Therefore, \(0.5 \text{ mol}\) will occupy \(\frac{22.4}{2} = 11.2 \text{ litres}\). Therefore, statement (a) is incorrect as it suggests 2.24 litres.

Step 3: Evaluate mol equivalent to N2

\( \mathrm{N}_2 \) has the same molar mass calculation logic: \(28 \text{ g/mol}\). Therefore, \(0.5 \text{ mol}\) of \(\mathrm{CO}\) is indeed equivalent in moles of \(\mathrm{N}_2\), making statement (c) correct.

Step 4: Check the number of molecules

Avogadro's Number states \(1 \text{ mol}\) contains \(6.02 \times 10^{23} \text{ molecules}\). Thus, \(0.5 \text{ mol}\) comprises \(0.5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23} \text{ molecules}\). Statement (d) is correct.

Key Concepts

Moles Calculation

To understand moles calculation, we must first be familiar with the term "mole". In chemistry, a mole is a standard unit of measurement for amount of substance. The number of moles of a substance corresponds to the number of constituent particles present.

To calculate the number of moles in a sample, you need to know its mass and molar mass. The formula to find moles is:

• Number of moles = \( \frac{\text{Mass of the substance}}{\text{Molar mass}} \)

The molar mass, given in grams per mole (g/mol), is the mass of one mole of a certain substance. For example, the molar mass of carbon monoxide (CO) is 28 g/mol, derived from carbon (12 g/mol) and oxygen (16 g/mol).

For a 14 g sample of CO, you’ll calculate moles using the formula:

• \( \frac{14 \text{ g}}{28 \text{ g/mol}} = 0.5 \text{ mol} \)

This shows that the sample contains 0.5 moles of CO.

Avogadro's Number

Avogadro's Number is a fundamental constant in chemistry that denotes the number of constituent particles, usually atoms or molecules, present in one mole of any substance. This number is approximately \(6.02 \times 10^{23}\).

Using Avogadro's Number, chemists can determine how many molecules or atoms are in a chemical sample. The formula to find the number of molecules in a given number of moles is:

• Number of molecules = Number of moles \( \times \) Avogadro's Number

For example, in our CO exercise, we have 0.5 moles of CO. To find how many molecules this represents:

• \( 0.5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23} \) molecules

Thus, 0.5 moles of CO comprises approximately 3.01 \(\times 10^{23}\) molecules, illustrating Avogadro's vital role in molecular calculations.

Gas Volume at NTP

The concept of gas volume at Normal Temperature and Pressure (NTP) is integral to understanding how gases behave under standard conditions. At NTP, which is 0 degrees Celsius and 1 atm pressure, one mole of any ideal gas occupies 22.4 liters.

For practical calculations, knowing this volume allows us to convert between moles and liters for a gas under these conditions. It's important to note how direct proportions work here:

• If we know the number of moles, we can find the volume by multiplying by the molar volume at NTP.
• Conversely, if we know the volume, we can determine the moles by dividing by 22.4 liters.

In our example with CO, we calculated that 0.5 moles of CO would occupy 11.2 liters, as 0.5 multiplied by 22.4 liters gives us this volume. This demonstrates that at NTP, half a mole of gas occupies exactly half the molar volume, reinforcing the utility of this principle in calculations.