Adding 1.56 g of K2SO4 to 6.00 mL of water at 16.2ºC causes the temperature of the solution to drop by 7.70ºC. How many grams of NaOH (ΔHsoln = –44.3 kJ/mol) would you need to add to raise the temperature back to 16.2ºC?

You need to add 0.243g of NaOH to raise the temperature back to 16.2°C

Explanation:

Using the equation:

Q = C*m*ΔT

Where Q is heat

C is specific heat

m is mass

and ΔT is change in temperature

We can find the heat required to increase the temperature of the solution back to 16.2°C:

Assuming specific heat of the solution of water + K2SO4 = Specific heat of water:

C = 4.184J/g°C

m = 1.56g + 6.00g = 7.56g

ΔT = 16.2°C - 7.70°C = 8.50°C

Q = 4.184J/g°C * 7.56g * 8.50°C

Q = 268.86J = 0.269kJ of heat are required

As this heat is obtained from the dissolution of NaOH:

0.269kJ * (1mol NaOH / 44.3kJ) = 0.00607 moles of NaOH are required

In grams -Molar mass NaOH: 40g/mol-:

0.00607 moles NaOH * (40g / mol) =