Suppose 15.0 grams of solid sodium hydroxide solid at 20.5 oC is dropped into 120 grams of water in a container that has a heat capacity of 10.0 J/oC, temperature also 20.5 oC. A solution that has a density of 1.04 g/mL forms at a temperature of 38.0 oC. What is the heat of solution of sodium hydroxide solid in water in kJ/mole ?
Specific Heats values for various solutions of sodium hydroxide
molarity NaOH (aq) 1.54 M, 2.89 M,4.16 M
Specific Heat 4.00 , 3.80 , 3.62
(J/goC)
To find the heat of solution of sodium hydroxide (NaOH) in water in kJ/mole, we can use the equation:
q = m * C * ΔT
where:
q = heat gained or lost by the solution (in joules)
m = mass of the solution (in grams)
C = specific heat capacity of the solution (in J/g·°C)
ΔT = change in temperature (in °C)
First, let's calculate the heat gained or lost by the water:
m_water = 120 g
C_water = 4.18 J/g·°C (specific heat capacity of water)
ΔT_water = 38.0°C - 20.5°C = 17.5°C
q_water = m_water * C_water * ΔT_water
Next, let's calculate the heat gained or lost by the sodium hydroxide:
m_NaOH = 15.0 g
C_NaOH = specific heat capacity of NaOH (we will use the value for 1.54 M NaOH)
ΔT_NaOH = 38.0°C - 20.5°C = 17.5°C
q_NaOH = m_NaOH * C_NaOH * ΔT_NaOH
Now, let's calculate the total heat gained or lost by the solution:
q_solution = q_water + q_NaOH
To convert the total heat gained or lost by the solution to kJ/mole, we need to know the number of moles of sodium hydroxide in the given mass.
To find the number of moles, we can use the atomic mass of NaOH:
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 39.00 g/mol
Number of moles = m_NaOH / Molar mass of NaOH
Finally, we can calculate the heat of solution of sodium hydroxide solid in water in kJ/mole:
Heat of solution = q_solution / Number of moles
By substituting the values into the equations and performing the calculations, you will find the heat of solution of sodium hydroxide in water in kJ/mole.