How much solution could be heated to boiling by the heat evolved by the dissolution of 26.5 g of naoh? (for the solution, assume a heat capacity of 4.0 j/g⋅∘c, an initial temperature of 25.0 ∘c, a boiling point of 100.0 ∘c, and a density of 1.05 g/ml.)

94.642 mL

Explanation:

Given:

Mass of the NaOH,  m = 26.5 g

Heat capacity of the solution, C = 4.0 J/g°C

Initial temperature = 25.0° C

Density = 1.05 g/mL

Now,

Molat mass of NaOH = 40 g/mol

Lattice energy of (NaOH) =  887 kJ/mol

The heat of hydration (NaOH) = 932 kJ/mol.

Thus,

The heat of dissolution of 1 mol of NaOH = 887 kJ/mol - 932 kJ/mol = -45 kJ/mol.

Number of moles of (NaOH), n = (26.5 g) / (40 g/mol) = 0.6625 mol.

Now, the change in temperature, ΔT = 100°C - 25°C = 75°C.

Thus,

the heat of dissolution of 0.6625 mol of NaOH, Q = -45 kJ/mol × 0.6625 mol = - 29.8125 kJ. = - 29812.5 J

Also

Q = m'CΔT

on substituting the values in the above formula, we get

- 29812.5 J = m' × 4 J/g·°C × 75°C.

m' = mass of the solution

or

m' = 99.375 g

m' = 90 g of solution.

Now,

Density = Mass / Volume

on substituting the values, we get

1.05g/mL = 99.375 g / Volume

or

Volume of the solution = 94.642 mL