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

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

Step 1: Understanding the Molar Mass of CO

The molar mass of carbon (C) is 12 g/mol, and the molar mass of oxygen (O) is 16 g/mol. Thus, the molar mass of carbon monoxide (CO) is calculated as the sum: \(12 + 16 = 28\) g/mol.

Step 2: Calculating Moles of CO

To find the number of moles in 14 g of CO, we use the formula \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). Substituting the values, we get \(\text{moles} = \frac{14}{28} = 0.5\) moles of CO.

Step 3: Mole-to-Volume Conversion at NTP

At normal temperature and pressure (NTP), 1 mole of any gas occupies 22.4 liters. Therefore, 0.5 moles of CO would occupy \(22.4 \times 0.5 = 11.2\) liters.

Step 4: Comparing Mole Quantities

The exercise states that 14 g of CO corresponds to the same moles as that of \(N_2\). However, since both CO and \(N_2\) have different molar masses, we need to verify the statement. CO is known to be 0.5 moles; for \(N_2\), 1 mole is 28 g (similar molar mass), hence 14 g would also be 0.5 moles.

Step 5: Calculating Number of Molecules

To find the number of molecules, we use Avogadro's number, which is approximately \(6.022 \times 10^{23}\) molecules/mol. For 0.5 moles of CO, the number of molecules would be \(0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}\) molecules.

Step 6: Identifying Incorrect Statement

Comparing all options, (a) is incorrect because 0.5 moles of CO should occupy 11.2 liters at NTP, not 2.24 liters. Other options are consistent with our calculations.

Key Concepts

Molar Mass Calculation

The concept of molar mass is crucial for solving chemistry problems. Molar mass is the mass of a single mole of a chemical element or compound. It is expressed in grams per mole (g/mol). Understanding how to calculate molar mass is key to navigating chemical equations and reactions.
For instance, to find the molar mass of carbon monoxide (CO), we add the atomic masses of carbon and oxygen. Carbon has an atomic mass of 12 g/mol, and oxygen has an atomic mass of 16 g/mol. Therefore, the molar mass of CO is the sum of these values:

• 12 g/mol (carbon)
• 16 g/mol (oxygen)

Thus, the molar mass of CO is 28 g/mol. This calculation helps us determine how much mass in grams one mole of CO would correspond to. When we know the mass of a compound sample, we can find out how many moles it represents by simply doing a division.

Mole-to-Volume Conversion

Mole-to-volume conversion is necessary when dealing with gases. At Normal Temperature and Pressure (NTP), which is defined as 0 degrees Celsius and 1 atmosphere pressure, one mole of any gas will occupy 22.4 liters. This is a universal value for gases at these conditions, also referred to as molar volume.
In practical situations, it is frequently necessary to convert the amount of substance in moles to the volume it occupies. For example, if we have 0.5 moles of carbon monoxide (CO), we can calculate its volume at NTP using the relationship:

• 1 mole of gas = 22.4 liters at NTP
• 0.5 moles of CO = 0.5 x 22.4 liters

This calculation results in 11.2 liters. Therefore, if you have half a mole of any gas under these conditions, it will occupy 11.2 liters. This conversion is fundamental for predicting how gases behave under specific conditions.

Avogadro's Number

Avogadro's number is an essential concept in chemistry that relates to counting particles on a molecular scale. It is defined as the number of constituent particles, usually atoms or molecules, in one mole of a given substance. The value is approximately 6.022 x 10^{23}. This large number helps bridge the gap between the atomic scale and the scale of grams we use in the lab.
For example, if you have 0.5 moles of carbon monoxide (CO), to find the number of molecules, you simply multiply the number of moles by Avogadro's number, like so:

• 0.5 moles x 6.022 x 10^{23} molecules/mole = 3.011 x 10^{23} molecules

This is how we can transition from counting moles to counting actual numbers of molecules, making Avogadro's number a cornerstone in the world of stoichiometry.

Gas Laws at NTP

The behavior of gases is often predictable using the gas laws, especially under standard conditions such as Normal Temperature and Pressure (NTP). At NTP, gases have certain properties that make calculations easier and more consistent. For instance, Boyle's Law and Charles's Law can be applied and simplified.
The concept of the molar volume, previously mentioned as being 22.4 liters per mole at NTP, is rooted in these laws. When problems involve gases at NTP, you can often skip complex calculations by applying this known molar volume, significantly streamlining the process.
Moreover, to check the accuracy of claims or conditions (like the ones in our exercise), knowing that these universal values and conditions apply can be a straightforward way to verify calculations. It offers a reliable foundation to check assumptions against actual chemical behavior.