Answer :
To determine how much energy is required to boil [tex]\( 100 \, \text{mL} \)[/tex] of water, we need to perform the following steps:
1. Convert the volume of water to mass: Given that the density of water is [tex]\( 1 \, \text{g/mL} \)[/tex], the mass [tex]\( m \)[/tex] of [tex]\( 100 \, \text{mL} \)[/tex] of water can be calculated as:
[tex]\[ m = 100 \, \text{mL} \times 1 \, \frac{\text{g}}{\text{mL}} = 100 \, \text{g} \][/tex]
2. Convert mass of water to moles: Using the molar mass of water, [tex]\( 18.02 \, \text{g/mol} \)[/tex], we can calculate the number of moles [tex]\( n \)[/tex] of water in [tex]\( 100 \, \text{g} \)[/tex]:
[tex]\[ n = \frac{100 \, \text{g}}{18.02 \, \text{g/mol}} \approx 5.55 \, \text{mol} \][/tex]
3. Calculate energy required for different constants:
- For Option A: The energy per mole is [tex]\( 4.186 \, \text{kJ/mol} \)[/tex], so the total energy [tex]\( E_A \)[/tex] is:
[tex]\[ E_A = n \times 4.186 \, \frac{\text{kJ}}{\text{mol}} \approx 5.55 \, \text{mol} \times 4.186 \, \frac{\text{kJ}}{\text{mol}} = 23.23 \, \text{kJ} \][/tex]
- For Option B: The energy per mole is [tex]\( 6.03 \, \text{kJ/mol} \)[/tex], so the total energy [tex]\( E_B \)[/tex] is:
[tex]\[ E_B = n \times 6.03 \, \frac{\text{kJ}}{\text{mol}} \approx 5.55 \, \text{mol} \times 6.03 \, \frac{\text{kJ}}{\text{mol}} = 33.46 \, \text{kJ} \][/tex]
- For Option C: The energy per mole is [tex]\( -285.83 \, \text{kJ/mol} \)[/tex], so the total energy [tex]\( E_C \)[/tex] is:
[tex]\[ E_C = n \times (-285.83) \, \frac{\text{kJ}}{\text{mol}} \approx 5.55 \, \text{mol} \times (-285.83) \, \frac{\text{kJ}}{\text{mol}} = -1586.18 \, \text{kJ} \][/tex]
- For Option D: The energy per mole is [tex]\( 40.65 \, \text{kJ/mol} \)[/tex], so the total energy [tex]\( E_D \)[/tex] is:
[tex]\[ E_D = n \times 40.65 \, \frac{\text{kJ}}{\text{mol}} \approx 5.55 \, \text{mol} \times 40.65 \, \frac{\text{kJ}}{\text{mol}} = 225.58 \, \text{kJ} \][/tex]
Based on the results of our calculations:
- [tex]\( E_A \approx 23.23 \, \text{kJ} \)[/tex]
- [tex]\( E_B \approx 33.46 \, \text{kJ} \)[/tex]
- [tex]\( E_C \approx -1586.18 \, \text{kJ} \)[/tex]
- [tex]\( E_D \approx 225.58 \, \text{kJ} \)[/tex]
The correct option, which closely matches our calculated energy required to boil [tex]\( 100 \, \text{mL} \)[/tex] of water, is:
D. [tex]$100 \, \text{mL} \times \frac{1 \, \text{g}}{1 \, \text{mL}} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 40.65 \, \text{kJ/mol} = 226 \, \text{kJ}$[/tex]
1. Convert the volume of water to mass: Given that the density of water is [tex]\( 1 \, \text{g/mL} \)[/tex], the mass [tex]\( m \)[/tex] of [tex]\( 100 \, \text{mL} \)[/tex] of water can be calculated as:
[tex]\[ m = 100 \, \text{mL} \times 1 \, \frac{\text{g}}{\text{mL}} = 100 \, \text{g} \][/tex]
2. Convert mass of water to moles: Using the molar mass of water, [tex]\( 18.02 \, \text{g/mol} \)[/tex], we can calculate the number of moles [tex]\( n \)[/tex] of water in [tex]\( 100 \, \text{g} \)[/tex]:
[tex]\[ n = \frac{100 \, \text{g}}{18.02 \, \text{g/mol}} \approx 5.55 \, \text{mol} \][/tex]
3. Calculate energy required for different constants:
- For Option A: The energy per mole is [tex]\( 4.186 \, \text{kJ/mol} \)[/tex], so the total energy [tex]\( E_A \)[/tex] is:
[tex]\[ E_A = n \times 4.186 \, \frac{\text{kJ}}{\text{mol}} \approx 5.55 \, \text{mol} \times 4.186 \, \frac{\text{kJ}}{\text{mol}} = 23.23 \, \text{kJ} \][/tex]
- For Option B: The energy per mole is [tex]\( 6.03 \, \text{kJ/mol} \)[/tex], so the total energy [tex]\( E_B \)[/tex] is:
[tex]\[ E_B = n \times 6.03 \, \frac{\text{kJ}}{\text{mol}} \approx 5.55 \, \text{mol} \times 6.03 \, \frac{\text{kJ}}{\text{mol}} = 33.46 \, \text{kJ} \][/tex]
- For Option C: The energy per mole is [tex]\( -285.83 \, \text{kJ/mol} \)[/tex], so the total energy [tex]\( E_C \)[/tex] is:
[tex]\[ E_C = n \times (-285.83) \, \frac{\text{kJ}}{\text{mol}} \approx 5.55 \, \text{mol} \times (-285.83) \, \frac{\text{kJ}}{\text{mol}} = -1586.18 \, \text{kJ} \][/tex]
- For Option D: The energy per mole is [tex]\( 40.65 \, \text{kJ/mol} \)[/tex], so the total energy [tex]\( E_D \)[/tex] is:
[tex]\[ E_D = n \times 40.65 \, \frac{\text{kJ}}{\text{mol}} \approx 5.55 \, \text{mol} \times 40.65 \, \frac{\text{kJ}}{\text{mol}} = 225.58 \, \text{kJ} \][/tex]
Based on the results of our calculations:
- [tex]\( E_A \approx 23.23 \, \text{kJ} \)[/tex]
- [tex]\( E_B \approx 33.46 \, \text{kJ} \)[/tex]
- [tex]\( E_C \approx -1586.18 \, \text{kJ} \)[/tex]
- [tex]\( E_D \approx 225.58 \, \text{kJ} \)[/tex]
The correct option, which closely matches our calculated energy required to boil [tex]\( 100 \, \text{mL} \)[/tex] of water, is:
D. [tex]$100 \, \text{mL} \times \frac{1 \, \text{g}}{1 \, \text{mL}} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 40.65 \, \text{kJ/mol} = 226 \, \text{kJ}$[/tex]
