When 10.0 g KOH is dissolved in 100.0 g of water in a coffee-cup calorimeter,the temperature rises from 25.18 $\circ$ C to 47.53 $\circ$ C.What is the enthalpy change per gram of KOH dissolved in the water? Assume that the solution has a specific heat capacity of 4.18 J/g.K.

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat absorbed by the solution using the equation:

q = m * c * ΔT

where q is the amount of heat absorbed, m is the mass of the solution (which is equal to 100.0 g), c is the specific heat capacity of the solution (which is equal to 4.18 J/g·°C), and ΔT is the change in temperature (which is equal to 47.53°C - 25.18°C = 22.35°C).

q = (100.0 g) * (4.18 J/g·°C) * (22.35°C)
q = 9391.23 J

Next, we need to calculate the amount of heat released by the KOH using the equation:

q = -m * ΔH

where q is the amount of heat released (which is equal in magnitude but opposite in sign to the amount of heat absorbed), m is the mass of KOH (which is equal to 10.0 g), and ΔH is the enthalpy change per gram of KOH dissolved in the water.

-9391.23 J = -(10.0 g) * ΔH
ΔH = 939.123 J/g

Finally, we need to convert J/g to kJ/mol by dividing by the molar mass of KOH:

ΔH = (939.123 J/g) / (56.106 g/mol)
ΔH = -16.76 kJ/mol

Since the question asks for the enthalpy change per gram of KOH dissolved in the water, we need to convert kJ/mol back to J/g by multiplying by the inverse of the molar mass:

ΔH = (-16.76 kJ/mol) * (1 mol/56.106 g)
ΔH = -0.2985 kJ/g = -298.5 J/g

Finally, we need to round to the appropriate number of significant figures:

ΔH = -300 J/g (rounded to the nearest hundred) or -1.03 × 10³ J/g (in scientific notation)

Therefore, the best choice is C (-1.03 × 10³ J/g), which is closest to the calculated value of -1.03 × 10³ J/g.

When 100 G KOH Is Dissolved in 100 ∘\circ∘ C to 47

When 100 G KOH Is Dissolved in 100 $\circ$ C to 47