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Sagot :
To determine how the enthalpy of fusion ([tex]\(\Delta H_{\text{fusion}}\)[/tex]) is used to calculate the energy released when a mass of liquid freezes, let's go through the process step by step.
1. Understand the Concept of Enthalpy of Fusion:
- Enthalpy of fusion ([tex]\(\Delta H_{\text{fusion}}\)[/tex]) is the amount of energy required to change a substance from a solid to a liquid at its melting point. Conversely, when a liquid freezes into a solid, it releases this amount of energy.
2. Relate Mass to Moles:
- The energy released when a certain mass of liquid freezes is dependent on the number of moles of the substance. To convert the mass of the liquid (in grams) to moles, we use the molar mass of the substance ([tex]\(M\)[/tex], in g/mol).
3. Formulate the Energy Released:
- The number of moles ([tex]\(n\)[/tex]) can be calculated as:
[tex]\[ n = \frac{\text{grams liquid}}{\text{molar mass}} \][/tex]
- The energy released ([tex]\(q\)[/tex]) when this mass of liquid freezes can be calculated using:
[tex]\[ q = \text{number of moles} \times \Delta H_{\text{fusion}} \][/tex]
- Replacing the number of moles in the equation, we get:
[tex]\[ q = \left( \frac{\text{grams liquid}}{\text{molar mass}} \right) \times \Delta H_{\text{fusion}} \][/tex]
- Simplifying further:
[tex]\[ q = \text{grams liquid} \times \left( \frac{1}{\text{molar mass}} \right) \times \Delta H_{\text{fusion}} \][/tex]
4. Compare with the Given Options:
- A. [tex]\(\text{Grams liquid} \times \frac{1}{\Delta H_{\text{fusion}}}\)[/tex] does not match because it suggests dividing by [tex]\(\Delta H_{\text{fusion}}\)[/tex] instead of multiplying.
- B. [tex]\(\text{Grams liquid} \times \frac{\text{mol}}{\text{g}} \times \frac{1}{\Delta H_{\text{fusion}}}\)[/tex] is incorrect as it implies division by [tex]\(\Delta H_{\text{fusion}}\)[/tex] instead of multiplication.
- C. [tex]\(\text{Grams liquid} \times \frac{\text{mol}}{\text{g}} \times \Delta H_{\text{fusion}}\)[/tex] matches as it correctly incorporates the conversion from grams to moles and multiplication by [tex]\(\Delta H_{\text{fusion}}\)[/tex].
- D. [tex]\(\text{Grams liquid} \times \Delta H_{\text{fusion}}\)[/tex] does not account for conversion from grams to moles.
Conclusion:
The correct method is to use:
[tex]\[ \text{Grams liquid} \times \frac{1}{\text{molar mass}} \times \Delta H_{\text{fusion}} \][/tex]
which is best represented by option C:
[tex]\[ \text{Grams liquid} \times \frac{\text{mol}}{\text{g}} \times \Delta H_{\text{fusion}} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{C} \][/tex]
1. Understand the Concept of Enthalpy of Fusion:
- Enthalpy of fusion ([tex]\(\Delta H_{\text{fusion}}\)[/tex]) is the amount of energy required to change a substance from a solid to a liquid at its melting point. Conversely, when a liquid freezes into a solid, it releases this amount of energy.
2. Relate Mass to Moles:
- The energy released when a certain mass of liquid freezes is dependent on the number of moles of the substance. To convert the mass of the liquid (in grams) to moles, we use the molar mass of the substance ([tex]\(M\)[/tex], in g/mol).
3. Formulate the Energy Released:
- The number of moles ([tex]\(n\)[/tex]) can be calculated as:
[tex]\[ n = \frac{\text{grams liquid}}{\text{molar mass}} \][/tex]
- The energy released ([tex]\(q\)[/tex]) when this mass of liquid freezes can be calculated using:
[tex]\[ q = \text{number of moles} \times \Delta H_{\text{fusion}} \][/tex]
- Replacing the number of moles in the equation, we get:
[tex]\[ q = \left( \frac{\text{grams liquid}}{\text{molar mass}} \right) \times \Delta H_{\text{fusion}} \][/tex]
- Simplifying further:
[tex]\[ q = \text{grams liquid} \times \left( \frac{1}{\text{molar mass}} \right) \times \Delta H_{\text{fusion}} \][/tex]
4. Compare with the Given Options:
- A. [tex]\(\text{Grams liquid} \times \frac{1}{\Delta H_{\text{fusion}}}\)[/tex] does not match because it suggests dividing by [tex]\(\Delta H_{\text{fusion}}\)[/tex] instead of multiplying.
- B. [tex]\(\text{Grams liquid} \times \frac{\text{mol}}{\text{g}} \times \frac{1}{\Delta H_{\text{fusion}}}\)[/tex] is incorrect as it implies division by [tex]\(\Delta H_{\text{fusion}}\)[/tex] instead of multiplication.
- C. [tex]\(\text{Grams liquid} \times \frac{\text{mol}}{\text{g}} \times \Delta H_{\text{fusion}}\)[/tex] matches as it correctly incorporates the conversion from grams to moles and multiplication by [tex]\(\Delta H_{\text{fusion}}\)[/tex].
- D. [tex]\(\text{Grams liquid} \times \Delta H_{\text{fusion}}\)[/tex] does not account for conversion from grams to moles.
Conclusion:
The correct method is to use:
[tex]\[ \text{Grams liquid} \times \frac{1}{\text{molar mass}} \times \Delta H_{\text{fusion}} \][/tex]
which is best represented by option C:
[tex]\[ \text{Grams liquid} \times \frac{\text{mol}}{\text{g}} \times \Delta H_{\text{fusion}} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{C} \][/tex]
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