A common flaw in calorimetry experiments to determine the molar enthalpy of solution is to have a mass of solute that is inappropriate for the volume of water used resulting in a temperature change that is too small or too large. A temperature change of 5°C to 10°C is a good range. Calculate the mass of NaOH ( molar enthalpy of the solution is – 44.51 kJ/mol ) that should be dissolved into 250 mL of water in order to produce a temperature increase of 7.00°C.

Question

A common flaw in calorimetry experiments to determine the molar enthalpy of solution is to have a mass of solute that is inappropriate for the volume of water used resulting in a temperature change that is too small or too large. A temperature change of 5°C to 10°C is a good range. Calculate the mass of NaOH ( molar enthalpy of the solution is – 44.51 kJ/mol ) that should be dissolved into 250 mL of water in order to produce a temperature increase of 7.00°C.

how would I go about and do this question?

heat= molesNaOH*molarenthalpy
but also
heat= masssolution*c*deltaTEMP

set these equal

MoleNaOH*44.51= volumewater*denstiywater*c*10C

solve for moleNaoh and convert to grams.

Answer 1

To calculate the mass of NaOH that should be dissolved into 250 mL of water to produce a temperature increase of 7.00°C, you can follow these steps:

1. Determine the heat absorbed by the solution using the molar enthalpy of the NaOH solution and the number of moles of NaOH.

heat = molesNaOH * molar enthalpy of solution

2. Calculate the heat absorbed by the solution using the mass of the solution, its specific heat capacity, and the change in temperature.

heat = mass solution * specific heat capacity * delta TEMP

3. Set these two expressions for heat equal to each other and solve for moles of NaOH.

molesNaOH * molar enthalpy of solution = mass solution * specific heat capacity * delta TEMP

4. Rearrange the equation to solve for moles of NaOH:

molesNaOH = (mass solution * specific heat capacity * delta TEMP) / molar enthalpy of solution

5. Convert moles of NaOH to grams by multiplying it by the molar mass of NaOH (which is 40 g/mol).

Let's plug in the given values:

Volumewater = 250 mL = 0.250 L (convert mL to liters)
Denstiywater = 1 g/mL (assume density of water is 1 g/mL)
C = specific heat capacity of water = 4.18 J/g·°C (common value for water)
Molar enthalpy of the solution = -44.51 kJ/mol (convert to J/mol by multiplying by 1000)
Delta TEMP = 7.00°C

First convert molar enthalpy from kJ/mol to J/mol:

molar enthalpy of solution = -44.51 kJ/mol * 1000 J/kJ = -44,510 J/mol

Now we can substitute these values into the equation:

molesNaOH = (mass solution * specific heat capacity * delta TEMP) / molar enthalpy of solution

molesNaOH = (0.250 L * 1 g/mL * 4.18 J/g·°C * 7.00°C) / -44,510 J/mol

Simplifying the equation:

molesNaOH = (0.250 * 4.18 * 7.00) / -44,510

molesNaOH = 0.0652 mol

Finally, convert moles of NaOH to grams:

massNaOH = molesNaOH * molar mass

massNaOH = 0.0652 mol * 40 g/mol

massNaOH = 2.608 g

Therefore, you would need to dissolve approximately 2.608 grams of NaOH into 250 mL of water to produce a temperature increase of 7.00°C.