Question

7.41 g ca(oh)₂ dissociates in 50.0 g of water in a calorimeter. the solution temperature increases from 21.50 °c to 28.25 °c. what is the heat of solution for the dissociation process?
ca(oh)₂ → ca²⁺ + 2oh⁻ cₛₒₗₙ = 4.18 j/g °c
mₛₒₗₙ = m_ca(oh)₂ + m_h₂o
qₛₒₗₙ = (mₛₒₗₙ)(cₛₒₗₙ)(δt)
qₛₒₗₙ = ? j
enter either a + or - sign and the magnitude.

Explanation:

Step1: Calculate mass of solution

First, find the mass of the solution by adding the mass of \( \text{Ca(OH)}_2 \) and water.
\( m_{\text{soln}} = m_{\text{Ca(OH)}_2} + m_{\text{H}_2\text{O}} = 7.41\ \text{g} + 50.0\ \text{g} = 57.41\ \text{g} \)

Step2: Calculate temperature change

Next, find the change in temperature \( \Delta T \).
\( \Delta T = T_{\text{final}} - T_{\text{initial}} = 28.25\ ^\circ\text{C} - 21.50\ ^\circ\text{C} = 6.75\ ^\circ\text{C} \)

Step3: Calculate heat absorbed by solution

Now, use the formula \( q_{\text{soln}} = (m_{\text{soln}})(c_{\text{soln}})(\Delta T) \) to find the heat.
Substitute the values: \( m_{\text{soln}} = 57.41\ \text{g} \), \( c_{\text{soln}} = 4.18\ \text{J/g}\ ^\circ\text{C} \), \( \Delta T = 6.75\ ^\circ\text{C} \)
\( q_{\text{soln}} = 57.41\ \text{g} \times 4.18\ \text{J/g}\ ^\circ\text{C} \times 6.75\ ^\circ\text{C} \)
First, multiply \( 57.41 \times 4.18 \approx 240.0 \) (approximate for simplicity, but we'll do exact calculation)
\( 57.41 \times 4.18 = 57.41 \times (4 + 0.18) = 57.41 \times 4 + 57.41 \times 0.18 = 229.64 + 10.3338 = 239.9738 \)
Then multiply by \( 6.75 \): \( 239.9738 \times 6.75 \approx 239.9738 \times 6 + 239.9738 \times 0.75 = 1439.8428 + 179.98035 = 1619.82315 \approx 1620\ \text{J} \) (more accurately, using calculator: \( 57.41 \times 4.18 \times 6.75 = 57.41 \times (4.18 \times 6.75) \))
\( 4.18 \times 6.75 = 28.215 \)
\( 57.41 \times 28.215 \approx 57.41 \times 28 + 57.41 \times 0.215 = 1607.48 + 12.34315 = 1619.82315 \approx 1620\ \text{J} \)
Since the temperature of the solution increased, the dissolution process is exothermic (releases heat), so the heat of solution (heat absorbed by solution is positive, but the heat of dissociation would be negative, but here we are calculating \( q_{\text{soln}} \) which is the heat absorbed by the solution, so it's positive? Wait, no: when the solution temperature increases, the process (dissociation) is releasing heat, so the system (dissociation) releases heat, so \( q_{\text{system}} = -q_{\text{surroundings}} \). But the question is asking for \( q_{\text{soln}} \), which is the heat gained by the solution, so it's positive. Wait, let's check the formula again. The formula given is \( q_{\text{soln}} = (m_{\text{soln}})(c_{\text{soln}})(\Delta T) \). Since \( \Delta T \) is positive (temperature increased), \( q_{\text{soln}} \) is positive, meaning the solution absorbed heat, so the dissociation process released heat (exothermic), so \( q_{\text{soln}} \) is positive. Wait, but let's do the calculation correctly.

Wait, let's recalculate:

\( m_{\text{soln}} = 7.41 + 50.0 = 57.41\ \text{g} \)

\( \Delta T = 28.25 - 21.50 = 6.75\ ^\circ\text{C} \)

\( q_{\text{soln}} = 57.41\ \text{g} \times 4.18\ \text{J/g}\ ^\circ\text{C} \times 6.75\ ^\circ\text{C} \)

Calculate step by step:

First, 57.41 * 4.18:

57.41 * 4 = 229.64

57.41 * 0.18 = 10.3338

Sum: 229.64 + 10.3338 = 239.9738

Then, 239.9738 * 6.75:

239.9738 * 6 = 1439.8428

239.9738 * 0.75 = 179.98035

Sum: 1439.8428 + 179.98035 = 1619.82315 ≈ 1620 J (rounded to three significant figures? Wait, the given values: 7.41 (three sig figs), 50.0 (three), 21.50 (four), 28.25 (four), 4.18 (three). So the least number of sig figs is three (from 7.41, 50.0, 4.18). Wait, 50.0 has three, 7.41 three, 4.18 three. So the answer should have three sig figs. Wait, 1619.82315 rounds to 1620 J? Wait, 1619.82 is approximately 1620 J (three sig figs: 1.62 x 10^3). But let's check the exact calculation:

57.41 * 4.18 = 57.41 * 4.18 = let's do 57.41 * 4.18:

57.41
x 4.18
--------
459.28 (57.41 * 0.08)
574.1 (57.41 * 0.1)
2296.4 (57.41 * 4)
--------
Wait, no, that's not the right way. Wait, 4.18 is 4 + 0.1 + 0.08. So 57.41 * 4 = 229.64, 57.41 * 0.1 = 5.741, 57.41 * 0.08 = 4.5928. Then sum: 229.64 + 5.741 = 235.381 + 4.5928 = 239.9738. Then multiply by 6.75: 239.9738 * 6.75. Let's do 239.9738 * 6 = 1439.8428, 239.9738 * 0.75 = 179.98035, total 1619.82315. So approximately 1620 J. Since the temperature increased, the solution absorbed heat, so \( q_{\text{soln}} \) is positive. Wait, but the question says "heat of solution for the dissociation process". Wait, maybe I made a mistake. Wait, the heat of solution is the heat change when the solute dissolves in the solvent. If the solution temperature increases, the process is exothermic, so the heat of solution (ΔH_soln) is negative. But the question is asking for \( q_{\text{soln}} \), which is the heat gained by the solution, so it's positive. Wait, the formula given is \( q_{\text{soln}} = (m_{\text{soln}})(c_{\text{soln}})(\Delta T) \), so we just need to calculate that. So:

\( q_{\text{soln}} = 57.41\ \text{g} \times 4.18\ \text{J/g}\ ^\circ\text{C} \times 6.75\ ^\circ\text{C} \approx 57.41 \times 4.18 \times 6.75 \approx 1620\ \text{J} \) (positive, since the solution absorbed heat). Wait, but let's check the significant figures. 7.41 (3), 50.0 (3), 4.18 (3), ΔT = 6.75 (3, since 28.25 - 21.50 = 6.75, which is three decimal places? Wait, 28.25 - 21.50 = 6.75, which is two decimal places, but the number of significant figures is three (6.75 has three). So the product should have three significant figures. 57.41 (four) * 4.18 (three) * 6.75 (three) = the least is three, so 1620 J (which is 1.62 x 10^3 J) or 1620 J. Wait, 1619.82315 rounds to 1620 J (three significant figures: 1.62 x 10^3). But let's see, maybe the question expects the exact value. Wait, let's do the calculation again:

57.41 * 4.18 = 239.9738

239.9738 * 6.75 = 239.9738 * 6.75

Let's compute 240 * 6.75 = 1620, so it's approximately 1620 J. Since the temperature increased, the solution gained heat, so \( q_{\text{soln}} \) is positive. So the answer is +1620 J (or more accurately, +1620 J when rounded to three significant figures). Wait, but let's check the exact calculation:

57.41 * 4.18 = 239.9738

239.9738 * 6.75 = 239.9738 * (6 + 0.75) = 239.9738*6 + 239.9738*0.75 = 1439.8428 + 179.98035 = 1619.82315 ≈ 1620 J (when rounded to three significant figures). So the heat of solution (q_soln) is +1620 J (or more precisely, +1620 J). Wait, but maybe the question considers that the dissociation is exothermic, so the heat of solution (the heat released by the dissociation) would be negative, but the formula given is for q_soln, which is the heat absorbed by the solution, so it's positive. Let's confirm with the formula:

Given \( q_{\text{soln}} = (m_{\text{soln}})(c_{\text{soln}})(\Delta T) \), with \( \Delta T = 6.75 \) (positive), so \( q_{\text{soln}} \) is positive. So the answer is +1620 J (or 1620 J with a + sign).

Answer:

+1620