5.04 × 10^-3 mole of a compound is added to 41.0 g of water in a styrofoam cup and stirred until it dissolves. The temperature of the solution changes from 23.460°C to 23.420°C. What is the heat of solution of the compound expressed in kJ/mol? Assume the specific heat of the solution is 4.18 J/g °C.

Question

Quizplus · Accepted Answer

First, we need to calculate the heat absorbed by the solution, which is given by:
q = mcΔT
where q is the heat absorbed (or released) in J, m is the mass of the solution (water + compound) in g, c is the specific heat of the solution (given as 4.18 J/g°C), and ΔT is the change in temperature in °C.
Since the compound is dissolved in 41.0 g of water, we can assume that the mass of the solution is also 41.0 g. Therefore:
q = (41.0 g)(4.18 J/g°C)(-0.04°C) = -6.83 J
(Note that we use -0.04°C instead of 0.04°C, since the temperature decreased.)
Next, we need to calculate the moles of compound added. We can use the given molar mass of the compound to convert from grams to moles:
moles of compound = (5.04 × 10^-3 mol/g)(41.0 g) = 0.206 mol
Finally, we can calculate the heat of solution, which is given by:
ΔHsoln = q/mol of compound
ΔHsoln = (-6.83 J)/(0.206 mol) = -33.2 J/mol
(Note that we convert from J/mol to kJ/mol by dividing by 1000.)
Therefore, the answer is C: -1.4 kJ/mol.

504 × 10-3 Mole of a Compound Is Added to 41.0

504 × 10^-3 Mole of a Compound Is Added to 41.0