Question

a student determines the heat of dissolution of solid calcium bromide using a coffee - cup calorimeter of negligible heat capacity. when 1.56 g of cabr₂(s) is dissolved in 102.00 g of water, the temperature of the solution increases from 25.00 to 26.95 °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). δh dissolution = kj/mol. use the references to access important values if needed for this question. 2 item attempts remaining. try another version submit answer

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. So $m = 1.56\ g+ 102.00\ g=103.56\ g$.

Step2: Calculate the heat absorbed by the solution

Use the formula $q = mc\Delta T$, where $c = 4.184\ J/(g\cdot^{\circ}C)$, $\Delta T=T_2 - T_1=26.95^{\circ}C - 25.00^{\circ}C = 1.95^{\circ}C$. Then $q=(103.56\ g)\times4.184\ J/(g\cdot^{\circ}C)\times1.95^{\circ}C$.

$$
LATEXBLOCK0
$$

Step3: Calculate the number of moles of $CaBr_2$

The molar - mass of $CaBr_2$: $M_{CaBr_2}=40.08\ g/mol+(2\times79.90\ g/mol)=40.08\ g/mol + 159.80\ g/mol=199.88\ g/mol$. The number of moles of $CaBr_2$, $n=\frac{m}{M}=\frac{1.56\ g}{199.88\ g/mol}=0.0078\ mol$.

Step4: Calculate the enthalpy of dissolution

The enthalpy of dissolution $\Delta H_{dissolution}=\frac{q}{n}$. Since $q = 844.09\ J$ and $n = 0.0078\ mol$, convert $q$ to $kJ$ ($q = 0.84409\ kJ$). Then $\Delta H_{dissolution}=\frac{0.84409\ kJ}{0.0078\ mol}\approx108.2\ kJ/mol$.

Answer:

$108.2$