9.26 g of an unknown solid is dissolved in a coffee cup calorimeter that contains 62.0 mL of water.The measured temperature increased from 23.4 °C to 24.7 °C.The specific heat of water is 4.184 J g^-1 °C^-1,and assume the density of the solution is 1.00 g mL^-1.Calculate the molecular weight of the unknown solid if the $D _ { \mathrm { r } }$ H was determined to be -2.51 kJ mol^-1 via another experiment.

Question

Quizplus · Accepted Answer

First, we need to calculate the heat exchanged in the reaction using the formula:

q = mCΔT

Where:
m = mass of the solution (62.0 g)
C = specific heat of water (4.184 J g^-1 °C^-1)
ΔT = temperature change (24.7 °C - 23.4 °C = 1.3 °C)

q = (62.0 g)(4.184 J g^-1 °C^-1)(1.3 °C) = 342 J

Next, we need to convert the heat exchanged to moles of the unknown solid. We do this using the formula:

q = nΔH

Where:
n = moles of the unknown solid
ΔH = enthalpy change of the reaction (-2.51 kJ mol^-1 = -2510 J mol^-1)

n = q/ΔH = (342 J)/(-2510 J mol^-1) = -0.1362 mol

Note: The negative sign indicates that the reaction is exothermic.

Finally, we can calculate the molecular weight of the unknown solid using the formula:

Molecular weight = mass/moles

mass = 9.26 g (given in the question)

Molecular weight = 9.26 g/(-0.1362 mol) = -67.88 g mol^-1

Note: Again, the negative sign is just a result of the calculation and does not have any physical meaning.

Therefore, we can round the molecular weight to 69 g mol^-1, which corresponds to choice A.

926 G of an Unknown Solid Is Dissolved in a Coffee