Question

When a student dissolves 2.50 g of LiCl in 100.0 mL of water (100.0 g) the temperature rises from 24.0 oC to 29.11oC. What is ∆H in KJ/mol for the dissolution of LiCl in water? Make sure you include the correct sign for ∆H and units (with a space between number and unit)

Answer 1

Answer:

$36.273\ \text{kJ/mol}$

Explanation:

m = Mass of LiCl = 2.5 g

M = Molar mass of LiCl = 42.394 g/mol

c = Specific heat of water = $4.186\ \text{J/g}^{\circ}\text{C}$

$\Delta T$ = Change in temperature = $29.11-24=5.11\ ^{\circ}\text{C}$

$m_w$ = Mass of water = $\rho V=1\times 100=100\ \text{g}$

Number of moles

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{2.5}{42.394}\\\Rightarrow n=0.05897\ \text{mol}$

Heat is given by

$Q=m_wc\Delta T\\\Rightarrow Q=100\times 4.186\times 5.11\\\Rightarrow Q=2139.046\ \text{J}$

Enthalpy is given by

$\Delta H=\dfrac{Q}{n}\\\Rightarrow \Delta H=\dfrac{2139.046}{0.05897}\\\Rightarrow \Delta H=36273.46\ \text{J/mol}=36.273\ \text{kJ/mol}$

The enthalpy for the dissolution is $36.273\ \text{kJ/mol}$.