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To determine the energy generated from freezing [tex]\(2.5 \text{ g}\)[/tex] of water, you need to follow these steps:
1. Determine the number of moles of water:
- Given the mass of water [tex]\(2.5 \text{ g}\)[/tex] and its molar mass [tex]\(18.02 \text{ g/mol}\)[/tex], calculate the number of moles using the formula:
[tex]\[ \text{moles of water} = \frac{\text{mass of water}}{\text{molar mass}} \][/tex]
Plugging in the values:
[tex]\[ \text{moles of water} = \frac{2.5 \text{ g}}{18.02 \text{ g/mol}} \approx 0.1387 \text{ mol} \][/tex]
2. Calculate the energy generated from freezing:
- The enthalpy change for the phase transition from liquid to solid (freezing point) is known as the enthalpy of fusion ([tex]\(\Delta H_{\text{fusion}}\)[/tex]), which is given as [tex]\(6.03 \text{ kJ/mol}\)[/tex].
- The energy generated can be calculated by multiplying the number of moles of water by the enthalpy of fusion:
[tex]\[ \text{energy generated} = \text{moles of water} \times \Delta H_{\text{fusion}} \][/tex]
Plugging in the values:
[tex]\[ \text{energy generated} = 0.1387 \text{ mol} \times 6.03 \text{ kJ/mol} \approx 0.8366 \text{ kJ} \][/tex]
Therefore, the correct option is:
B. [tex]\(2.5 \text{ g} \times \frac{1 \text{ mol}}{18.02 \text{ g}} \times 6.03 \text{ kJ/mol}\)[/tex]
1. Determine the number of moles of water:
- Given the mass of water [tex]\(2.5 \text{ g}\)[/tex] and its molar mass [tex]\(18.02 \text{ g/mol}\)[/tex], calculate the number of moles using the formula:
[tex]\[ \text{moles of water} = \frac{\text{mass of water}}{\text{molar mass}} \][/tex]
Plugging in the values:
[tex]\[ \text{moles of water} = \frac{2.5 \text{ g}}{18.02 \text{ g/mol}} \approx 0.1387 \text{ mol} \][/tex]
2. Calculate the energy generated from freezing:
- The enthalpy change for the phase transition from liquid to solid (freezing point) is known as the enthalpy of fusion ([tex]\(\Delta H_{\text{fusion}}\)[/tex]), which is given as [tex]\(6.03 \text{ kJ/mol}\)[/tex].
- The energy generated can be calculated by multiplying the number of moles of water by the enthalpy of fusion:
[tex]\[ \text{energy generated} = \text{moles of water} \times \Delta H_{\text{fusion}} \][/tex]
Plugging in the values:
[tex]\[ \text{energy generated} = 0.1387 \text{ mol} \times 6.03 \text{ kJ/mol} \approx 0.8366 \text{ kJ} \][/tex]
Therefore, the correct option is:
B. [tex]\(2.5 \text{ g} \times \frac{1 \text{ mol}}{18.02 \text{ g}} \times 6.03 \text{ kJ/mol}\)[/tex]
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