Question

A coffee cup calorimeter initially contains 125g water at $24.2°C$. Potassium bromide (10.5g), also at $\mathbf{24}\mathbf{.}\mathbf{2}\mathbf{°}\mathit{C}$, is added to the water, and after the KBr dissolves, the final temperature is$\mathbf{21}\mathbf{.}\mathbf{1}\mathbf{°}\mathit{C}$ . Calculate the enthalpy change for dissolving the salt in$J/gandkJ/mol$ . Assume that the specific heat capacity of the solution is$4.18J/°Cg$ and that no heat is transferred to the surroundings or to the calorimeter.

Short answer

The enthalpy change for dissolving the salt is$18.41kJ/mol$ . $∆H$is known as enthalpy change.

Step-by-step solution

Explanation

Heat,$q=mC∆T$

Now, $q=125g\times 4.18J/°Cg\times (21.1-24.28°C)=-1619.75J$

So the value of $q=-1619.75J$(Negative sign indicates that heat is absorbed).

Enthalpy change,$H=\frac{-1619.75J}{10.5g}=154.26J/g$

$NumberofMolesofKBr=\frac{MassofKBr}{MolecularmassofKBr}=\frac{10.5g}{119g/mol}=0.088Mole$

$∆H=1619.75/10.5g=154.26J/g$

$NumberofmolesofKBr=massofKBr/molecularweightofKBr=10.5g/119gmo{l}^{-1}$

 Step 2: To calculate  ∆H  

$$\begin{gathered} ∆H=\frac{-1619.75J}{0.088mole}=18406J/mol \\ =18.41kJ/mol \end{gathered}$$

$$\begin{gathered} ∆H=1619.75J/0.088mol \\ =18.41KJ/mol \end{gathered}$$