MCAT General Chemistry Question 263: Answer and Explanation

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Question: 263

5. The transfer of heat to or from a solution changes the temperature of the solution according to the equation q = mc∆T where q is the heat transferred, m is the mass of solvent, and c is the specific heat of the solvent. If a 1 g sample of a salt was dissolved in 20 mL of water (specific heat = 4.18 J/g°C) in an insulated beaker and the temperature was found to decrease by 4°C, which of the following salts was used? Assume no phase change for the water.

A. LiCl
B. KCH₃CO₂
C. NH₄NO₃
D. NaCl

Correct Answer: C

Explanation:

C Since the question states that the temperature of the solution decreased, the salts with exothermic dissolution enthalpies (choices A and B) can be eliminated. Using the calorimetry equation given in the question stem (q = mc∆T), we can estimate:

20 g × ≈ 4 J/g°C × 4°C = ≈ 320 J = ≈ 0.32 kJ = q

Since this heat is associated with 1 g of salt, in order to compare to the ∆H^o_diss in Table 2, convert this energy to a per mole basis by multiplying by the molar mass of the salt.

For choice C (NH₄NO₃, MW = 80 g/mol), this yields:

≈ 0.3 kJ/1g NH₄NO₃ × 80 g NH₄NO₃/mol = 24 kJ/mol

which is close to the given 25.69 kJ/mol in the table. The comparable calculation for NaCl yields:

≈ 0.3 kJ/1g NaCl × ≈ 60 g NaCl/mol = 18 kJ/mol

so choice D can be eliminated.

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