4. Below are data from a calorimetric determination of AHwl for an unknown salt.

Question

4. Below are data from a calorimetric determination of AHwl for an unknown salt. (Refer to part C of the procedure). Caleulate the enthalpy of solution per gram of salt, assuming that CPed is entirely attributable to the water (as in question 2). Identify the salt from the list in the above table. Why is Tr obtained by extrapolating temperature to time0 s? 8.79 g 110.07 g mass of calorimeter: mass ofcalbrimeterwater mass of calorimeter +water + salt: initial calorimeter temperature: final calorimeter temperature: 115.67 g 21.59±0.05 (see Fig. 2) 18.46 ± 0.05-C T,-21.59 ± aDS'C 20,5 Figure 2. Representative data for pre-lab assignment question #4

Explanation / Answer

Step 1. Calculate amount of heat released or absorbed.

Data from experiment:

Mass of water = 110.07 g – 8.79 = 101.28 g

cg = specific heat capacity of water = 4.18 J°C-1g-1

T = change in temperature

= |Tfinal - Tinitial|
= |21.59 - 18.46|
= 3.13°C

Calculate q (energy absorbed)

q = m × cg × T

q = 101.28 g × 4.18 J°C-1g-1 × 3.13°C = 1325.087 J

Note: temperature falls, so heat is absorbed.

Step 2. Calculate the enthalpy of dissolving per gram of salt

Mass of salt = 115.67 – 110.07 = 5.6 g

Heat change of dissolving 5.6 g of salt = 1325.087 J

Therefore Heat change of dissolving 1.0 g of salt = 1325.087 J/ 5.6 g = 236.622 J/g

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