Question

CaCl₂(s) → Ca²⁺(aq) + 2Cl⁻(aq) ∆H = -81.5 kJ/mol A 15.0-g sample of CaCl₂ is dissolved in 105 g water with both substances at 25.0°C. Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/°C･g.

Answer 1

Answer:

The final temperature of the solution is 46.93°C

Explanation:

we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = 81.5 kJ/mol

q = heat released = ?  

m = mass of CaCl_2 = 15.0 g

Molar mass of CaCl_2 = 110.98 g/mol

$\text{Moles of }CaCl_2\\\\=\frac{\text{Mass of }CaCl_2}{\text{Molar mass of }CaCl_2}\\=\frac{15.0g}{110.98g/mole}\\=0.135mole$

Now put all the given values in the above formula, we get:

$81.5kJ/mol=\frac{q}{0.135mole}q=11.0kJ$

calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 11.0 kJ = 11000 J

m = mass of solution = 15.0 + 105 = 120.0 g

c = specific heat capacity of water = 4.18J/g°C

$T_1$ = initial temperature = 25.0°C

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$11000J=120.0g\times 4.18J/g^oC\times (T_2-25.0)T_2=46.93^oC$

The final temperature of the solution is 46.93°C