Question

a student determines the heat of dissolution of solid calcium bromide using a coffee - cup calorimeter of negligible heat capacity. when 1.77 g of cabr2(s) is dissolved in 115.00 g of water, the temperature of the solution increases from 25.00 to 27.00 °c. based on the students observation, calculate the enthalpy of dissolution of cabr2(s) in kj/mol. assume the specific heat of the solution is 4.184 j/(g·°c). δhdissolution = kj/mol 3 item attempts remaining use the references to access important values if needed for this question.

Explanation:

Step1: Calculate the mass of the solution

The mass of the solution $m$ is the sum of the mass of $CaBr_2$ and the mass of water. $m = 1.77\ g+ 115.00\ g=116.77\ g$.

Step2: Calculate the heat absorbed by the solution

Use the formula $q = mc\Delta T$, where $c = 4.184\ J/(g\cdot^{\circ}C)$, $\Delta T=T_{final}-T_{initial}=27.00^{\circ}C - 25.00^{\circ}C = 2.00^{\circ}C$. Then $q=(116.77\ g)\times4.184\ J/(g\cdot^{\circ}C)\times2.00^{\circ}C$.
$q = 116.77\times4.184\times2.00\ J\approx978.4\ J$.

Step3: Calculate the number of moles of $CaBr_2$

The molar - mass of $CaBr_2$ is $M = 40.08\ g/mol+2\times79.90\ g/mol=40.08\ g/mol + 159.8\ g/mol=199.88\ g/mol$. The number of moles $n=\frac{m}{M}$, where $m = 1.77\ g$. So $n=\frac{1.77\ g}{199.88\ g/mol}\approx0.008855\ mol$.

Step4: Calculate the enthalpy of dissolution

The enthalpy of dissolution $\Delta H_{dissolution}=\frac{q}{n}$. Since $q = 978.4\ J$ and $n = 0.008855\ mol$, convert $q$ to $kJ$ ($q = 0.9784\ kJ$). Then $\Delta H_{dissolution}=\frac{0.9784\ kJ}{0.008855\ mol}\approx110.5\ kJ/mol$.

Answer:

$110.5$