Q4. Using ΔsolH = −73.1 kJ mol−1, predict the temperature change when 3.00 g of anhydrous CuSO4 (M = 159.6 g mol−1) dissolves in 100 mL of water in a calorimeter with W = 34.9 J K−1.
Moles = 3.00/159.6, about 0.01880 mol, so the heat given out is about 0.01880 × 73 100, or 1374 J. Heat capacity = 103 × 4.184 + 34.9, about 465.9 J K−1. Δt = 1374/465.9: a rise of about 2.9 °C.
Q5. Why is the mass of the solution taken as (100 + mass of salt) g and not just 100 g?
After dissolving, the salt is part of the solution, and the whole solution changes temperature. Its heat capacity is therefore that of 100 g of water plus the salt, taking its specific heat as water’s, 4.184 J g−1 K−1.
Q6. The solubility of copper sulphate crystals in water rises with temperature. How does this fit the sign of their enthalpy of solution? (Adapted from discussion question (v), NCERT Laboratory Manual, page 24.)
Dissolving the crystals absorbs heat (ΔsolH positive). For an endothermic change, raising the temperature shifts the balance towards dissolving, so more salt dissolves in hot water, as NCERT notes for most salts with positive ΔsolH (page 156).
Q7. While the crystals dissolve, the solution is colder than the room. How does this affect the measured ΔsolH, and how can the error be reduced?
Heat leaks in from the room, so the solution does not cool as much as it should; the measured fall, and so ΔsolH, comes out too small. Insulate well, dissolve the powder quickly with steady stirring, take the lowest reading, or draw the slow warming line back to the moment the salt was added and read the fall there.
Q8. A student uses the readings of Example 1 (heat taken in 118.6 J for 5.00 g of crystals) but divides by the molar mass of the anhydrous salt, 159.6 g mol−1. What does he report, and is it too large or too small?
Moles = 5.00/159.6, about 0.03133 mol, so he reports 118.6/0.03133 J mol−1, about +3.8 kJ mol−1. It is too small: he has counted more moles than were dissolved; the correct value is about +5.9 kJ mol−1.