When a 7.50 g sample of solid NaOH dissolves in100.00 g of water in a coffee-cup calorimeter, the temperature rises from 21.6°C to 37.8°C. Calculate the ∆H (in kJ/mol NaOH) for the solution process. (Assume the specific heat of the solution formed is 4.18 J/g-K). Note: The total mass= 107.50 g. (Atomic mass of O-15.999 g/mol; H-1.008 g/mol; Na-22.990 g/mol).

NaOH(s) → Na+(aq) + OH– (aq)

-5.44 kJ/mol

-44.30 kJ/mol

-38.83 kJ/mol

-2.67- kJ/mol

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First, calculate the heat absorbed by the solution using the formula:

q = mCΔT

where q is the heat absorbed (in Joules), m is the mass of the solution (in grams), C is the specific heat of the solution (in J/g-K), and ΔT is the change in temperature (in K).

m of solution = 100.00 g + 7.50 g = 107.50 g

ΔT = 37.8°C - 21.6°C = 16.2 K

q = (107.50 g)(4.18 J/g-K)(16.2 K) = 7273 J

Next, convert the heat absorbed to the heat of solution (in kJ/mol):

-7273 J x (1 kJ/1000 J) x (1 mol NaOH/7.50 g NaOH) = -0.97 kJ/mol NaOH

Finally, since the reaction involves the dissolution of one mole of NaOH, the ∆H (in kJ/mol NaOH) for the solution process is equal to -0.97 kJ/mol NaOH.

The closest answer choice is -2.67 kJ/mol, which is not the exact answer calculated but may be the result of a rounding error during the calculation process.

To calculate the ∆H (in kJ/mol NaOH) for the solution process, we can use the equation:

∆H = q / n

Where:
∆H = enthalpy change
q = heat absorbed or released by the solution
n = number of moles of NaOH

First, let's calculate the heat absorbed or released by the solution, q.

q = m × c × ∆T

Where:
m = mass of the solution = mass of water + mass of NaOH
c = specific heat of the solution
∆T = change in temperature = final temperature - initial temperature

mass of water = 100.00 g
mass of NaOH = 7.50 g
mass of the solution = 100.00 g + 7.50 g = 107.50 g

∆T = 37.8°C - 21.6°C = 16.2°C

Now, let's convert the temperature to Kelvin:

∆T = 16.2°C + 273.15 = 289.35 K

Next, let's substitute the values into the equation:

q = 107.50 g × 4.18 J/g-K × 289.35 K

q = 130,132.065 J

Next, let's convert the heat into kJ:

q = 130,132.065 J ÷ 1000 = 130.132065 kJ

Now, let's calculate the number of moles of NaOH:

moles of NaOH = mass of NaOH / molar mass of NaOH

molar mass of NaOH = 22.990 g/mol + 15.999 g/mol + 1.008 g/mol = 39.997 g/mol

moles of NaOH = 7.50 g / 39.997 g/mol

moles of NaOH ≈ 0.1875 mol

Now, let's substitute the values into the equation:

∆H = 130.132065 kJ / 0.1875 mol

∆H ≈ -694.3746667 kJ/mol

Rounding to the appropriate number of significant figures, we get:

∆H ≈ -694.4 kJ/mol

Therefore, the ∆H (in kJ/mol NaOH) for the solution process is approximately -694.4 kJ/mol.