Question

Open the Endothermic and Exothermic Reactions Lab Activity (Section 6.5, Figure 6.9) to complete this worksheet. 1. Measure the temperature change after dissolving a 5 g sample of each substance in 100 mL of water. Determine which solutes have an exothermic dissolving process, and which have an endothermic dissolving process. Naci NaOH NaNO3 CaCl2 KOH NH4NO3 Temperature change (°C) 1-0.79 13.30 -2.88 | Exo- or Endo-? | exo endo exo 8.92 12.27 -3.84 endo endo exo 2. Make a prediction about how these values will change when doubling the amount of solute, water, or both. Then test your predictions. NaOH NaNO3 CaCl2 KOH NH4NO3 Temperature NaCl change (°C) 10 g in 100 mL-1.59 5 g in 200 mL -0.40 10 g in 200 mL -0.79 26.60 -5.7617.83 44.54 -7.67 6.65 -1.44 4:46.6.14 -1.92 13.30 -2.88 8.92 12.27 -3.84 3. How much heat energy is released by dissolving 10.0 g of CaCl2 in 100.0 mL of water? Use the formula q=mcAT. The specific heat of water is 4.184 J/g. °C, and the density of water is 1.00 g/mL.

(#5)

How much heat is absorbed by dissolving 10.0 g of NaNO3 in 100.0 mL of water?

Answer 1

5. Mass of water used=Volume of water x density of water

=100 mL x 1 g/mL=100 g

Mass of solution=Mass of NaNO​​​3 + mass of water

=10 g + 100 g=110 g

From table 1, fall in temperature of solution on adding 5 g NaNO​​​3=-2.88°C (this means the dissolution of NaNO​​​3 absorbs heat from the solution resulting in fall of temperature of solution)

So we predict (as given in table 2), fall in temperature of solution on adding 10 g (2 x 5g) NaNO​​​3=-2.88°C x 2= -5.76°C

Given that we have to use specific heat of water c=4.184 J/g°C

To calculate Heat absorbed by dissolving 10.0 g NaNO​​​3 in 100 mL water

Heat released from solution=Mass of solution x specific heat of solution x fall in temperature of solution

=110 g x 4.184 J/g°C x (-5.76 °C)

=-2650.98 J

So heat absorbed by dissolution of 10 g NaNO3 in solution=+2650.98 J