Question

The enthalpy of solution (∆h) of nano₃ is 20.4 kj/mol. if 5.25 g nano₃ is dissolved in enough water to make a 100.0 ml solution, what is the change in temperature (°c) of the solution? (the specific heat capacity of the solution is 4.184 j/g･°c and the density of the solution is 1.02 g/ml).

Answer 1

The change in the temperature (°C) of the solution when 5.25 g of NaNO₃ is dissolved in enough water to make a 100.0 ml solution is 2.95 °C

How to determine the mole of NaNO₃

• Mass of NaNO₃ = 5.25 g
• Molar mass of NaNO₃ = 85 g/mol
• Mole of NaNO₃ =?

Mole = mass / molar mass

Mole of NaNO₃ = 5.25 / 85

Mole of NaNO₃ = 0.0618 mole

How to determine the heat

• Mole of NaNO₃ = 0.0618 mole
• Enthalpy change (ΔH) = 20.4 KJ/mol
• Heat (Q) =?

Q = n × ΔH

Q = 0.0618 × 20.4

Q = 1.26072 KJ

Q = 1.26072 × 1000

Q = 1260.72 J

How to determine the mass of the solution

• Density of solution = 1.02 g/mL
• Volume of solution = 100 mL
• Mass of solution =?

Mass = Density ×Volume

Mass of solution = 1.02 × 100

Mass of solution = 102 g

How to determine the change in the temperature

• Mass of solution (M) = 102 g
• Specific heat capacity (C) = 4.184 J/gºC
• Heat (Q) = 1260.72 J

• Change in temperature (ΔT) =?

Q = MCΔT

1260.72 = 102 × 4.184 × ΔT

1260.72 = 426.768 × ΔT

Divide both side by 426.768

ΔT = 1260.72 / 426.768

ΔT = 2.95 °C

Learn more about heat transfer:

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