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

`36.273\ \text{kJ/mol}`

Step-by-step 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* 100=100\ \text{g}`

Number of moles

`n=(m)/(M)\\\Rightarrow n=(2.5)/(42.394)\\\Rightarrow n=0.05897\ \text{mol}`

Heat is given by

`Q=m_wc\Delta T\\\Rightarrow Q=100* 4.186* 5.11\\\Rightarrow Q=2139.046\ \text{J}`

Enthalpy is given by

`\Delta H=(Q)/(n)\\\Rightarrow \Delta H=(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}`.