Answer:
ΔH per mole of CaCl2 is 160.3 kJ
Explanation:
<u>Step 1: </u>The balanced equation
CaCl2 (aq)→ Ca2+ + 2Cl-(aq)
<u>Step 2:</u> Data given
Molar mass of CaCl2 =110.98 g/mole
mass of CaCl2 = 4.02 grams
volume of the solution = 349 mL = 0.349 L
Temperature increases with 3.91 °C
Specific heat of the solution = specific hea of water = 4.18 J/°C*g
Density solution = density water = 1g/cm³
<u>Step 3:</u> Calculate moles of CaCl2
Numer of moles of CaCl2 = mass of CaCl2 / Molar mass of CaCl2
Number of moles = 4.02 grams /110.98 g/mole= 0.036 moles
<u>Step 4</u> = Calculate amount of heat added
Q = m*c*ΔT
with m = the mass 349 grams + 4.02 grams = 353.02 grams
c = 4.18 J/°C*g
ΔT = 3.91 °C
Q = 353.02*4.18*3.91 = 5769.69 J
Step 5: Calculate ΔH per mole
5769.69 J /0.036 moles = 160269.1 J = 160.3 kJ
ΔH per mole of CaCl2 is 160.3 kJ
