Which one of the following would weigh the most?

A. 1 mole of [tex]CO_2[/tex]
B. 1 mole of [tex]H_2O[/tex]
C. 1 mole of [tex]NH_3[/tex]
D. 1 mole of [tex]NO_2[/tex]

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The volume occupied by [tex]7.1 \, \text{g}[/tex] of chlorine gas at STP is:

A. [tex]22.4 \, \text{L}[/tex]
B. [tex]2.24 \, \text{L}[/tex]
C. [tex]11.2 \, \text{L}[/tex]
D. [tex]1.12 \, \text{L}[/tex]



Answer :

Certainly! Let's tackle each part of the question one at a time.

### Part i: Determining the heaviest molecule

To find out which one weighs the most, we will calculate the molar mass (molecular weight) of each substance (CO₂, H₂O, NH₃, NO₂). The molar mass of a molecule is the sum of the atomic masses of all the atoms in the molecule, given in grams per mole (g/mol).

1. CO₂ (Carbon dioxide):
- Carbon (C): atomic mass ≈ 12 g/mol
- Oxygen (O): atomic mass ≈ 16 g/mol
- Molar mass of CO₂ = 12 + 2 16 = 12 + 32 = 44 g/mol

2. H₂O (Water):
- Hydrogen (H): atomic mass ≈ 1 g/mol
- Oxygen (O): atomic mass ≈ 16 g/mol
- Molar mass of H₂O = 2
1 + 16 = 2 + 16 = 18 g/mol

3. NH₃ (Ammonia):
- Nitrogen (N): atomic mass ≈ 14 g/mol
- Hydrogen (H): atomic mass ≈ 1 g/mol
- Molar mass of NH₃ = 14 + 3 1 = 14 + 3 = 17 g/mol

4. NO₂ (Nitrogen dioxide):
- Nitrogen (N): atomic mass ≈ 14 g/mol
- Oxygen (O): atomic mass ≈ 16 g/mol
- Molar mass of NO₂ = 14 + 2
16 = 14 + 32 = 46 g/mol

Now, let’s compare the molar masses:
- CO₂: 44 g/mol
- H₂O: 18 g/mol
- NH₃: 17 g/mol
- NO₂: 46 g/mol

The heaviest molecule is NO₂ with a molar mass of 46 g/mol.

So, the answer is:
D. 1 mole of \( NO_2 \)

### Part ii: Volume occupied by 7.1 grams of chlorine gas at STP

At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters.
First, we need to find the number of moles of chlorine gas (Cl₂) in 7.1 grams.

1. Chlorine gas (Cl₂):
- Chlorine (Cl): atomic mass ≈ 35.5 g/mol
- Molar mass of Cl₂ = 2 * 35.5 = 71 g/mol

To calculate the number of moles:
[tex]\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{7.1 \text{ g}}{71 \text{ g/mol}} = 0.1 \text{ moles} \][/tex]

Now, we use the fact that 1 mole of gas occupies 22.4 liters at STP:
[tex]\[ \text{Volume of 0.1 moles} = 0.1 \times 22.4 \text{ L} = 2.24 \text{ L} \][/tex]

So, the volume occupied by 7.1 g of chlorine gas at STP is:
B. [tex]\( 2.24 \, \text{L} \)[/tex]