Answer
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Hint: Heat of solution is also called as the enthalpy of solution, which is the enthalpy change in relation with the dissolution of a solute in a solvent at a constant pressure, that results in infinite dilution. It has a unit of KJ/mol.
Step by step answer:
- We know that the heat energy released is equal to
$\begin{align}
& 125\times (24.2-21.1) \\
& =387.5\text{ }cal \\
\end{align}$
- Now the heat of solution =$\dfrac{heat\text{ }energy\text{ }released}{weight\text{ }of\text{ }solute}$
\[\dfrac{420.05}{10.5}ca\operatorname{l}\text{ }{{g}^{-1}}\]
=40cal/g
- So, one calories are equal to 4.126J,
Hence, heat of solution will be
\[\begin{align}
& 40\times 4.126\text{ }J/g \\
& =167\text{ }J/g \\
\end{align}\]
- There, we can conclude that the correct option is (c), that is the heat of solution is 167.7 J/g.
Additional information:
- Depending on the energy required to break the bonds initially and also about how much energy is released on bond formation between solute and solvent, the heat of solution can be endothermic or can be exothermic.
- It is also found that the heat of solution depends upon dissolution of solute in solvent at constant pressure, resulting in infinite dilution.
- We can say that the heat of solution is the sum of the enthalpy changes of the following three steps:
- The first step is the breaking of bonds within the solute, the next step is the breaking of the intermolecular interaction forces within the solvent and formation of new attractive solute-solute bonds.
Note:
- If more energy is released in making bonds than that used in breaking the bonds, then we can say that the overall process is exothermic. If more energy is used in breaking bonds than that is released on solute-solvent bond formation, then the overall process is endothermic.
Step by step answer:
- We know that the heat energy released is equal to
$\begin{align}
& 125\times (24.2-21.1) \\
& =387.5\text{ }cal \\
\end{align}$
- Now the heat of solution =$\dfrac{heat\text{ }energy\text{ }released}{weight\text{ }of\text{ }solute}$
\[\dfrac{420.05}{10.5}ca\operatorname{l}\text{ }{{g}^{-1}}\]
=40cal/g
- So, one calories are equal to 4.126J,
Hence, heat of solution will be
\[\begin{align}
& 40\times 4.126\text{ }J/g \\
& =167\text{ }J/g \\
\end{align}\]
- There, we can conclude that the correct option is (c), that is the heat of solution is 167.7 J/g.
Additional information:
- Depending on the energy required to break the bonds initially and also about how much energy is released on bond formation between solute and solvent, the heat of solution can be endothermic or can be exothermic.
- It is also found that the heat of solution depends upon dissolution of solute in solvent at constant pressure, resulting in infinite dilution.
- We can say that the heat of solution is the sum of the enthalpy changes of the following three steps:
- The first step is the breaking of bonds within the solute, the next step is the breaking of the intermolecular interaction forces within the solvent and formation of new attractive solute-solute bonds.
Note:
- If more energy is released in making bonds than that used in breaking the bonds, then we can say that the overall process is exothermic. If more energy is used in breaking bonds than that is released on solute-solvent bond formation, then the overall process is endothermic.
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