Question

5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the dissolution of MgSO₄. (The specific heat of water is 4.18 J/g･°C and the density of the water is 1.00 g/mL). You can assume that the specific heat of the solution is the same as that of water.

Answer 1

Answer: Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

Explanation:

To calculate the entalpy, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed by water = ?

m = mass of water = ${\text {volume of water}}\times {\text {density of water}}=100.0ml\times 1.00g/ml=100.0g$

c = heat capacity of water = 4.186 J/g°C

$\Delta T$= change in temperature = $6.70^0C$

$q=100.0g\times 4.184J/g^0C\times 6.70^0C=2803.3J=2.8033kJ$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

The heat absorbed by water will be equal to heat released by $MgSO_4$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 5.11 g

Molar mass  = 120 g/mol

Putting values in above equation, we get:

$\text{Moles of }MgSO_4=\frac{5.11g}{120g/mol}=0.042mol$

0.042 moles of $MgSO_4$ releases = 2.8033 kJ

1 mole of $MgSO_4$ releases = $\frac{2.8033 kJ}{0.042}\times 1=66.7kJ$

Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ