Answer to Question #135958 in General Chemistry for danielle

Question

A laboratory technician adds 43.1mL concentrated, 11.6 mol/ L hydrochloric acid to water to form 500.0mL of dilute solution. The temperature of the solution changes from 19.2. degrees celsius to 21.8 degrees celsius. Calculate the molar enthalpy of dilution of hydrochloric acid.

Answer 1

$43.1$mL solution$\times$ $($ $1L\over1000mL$ $)$ $\times$ $($ $11.6$ moles$HCl\over L$ Solution$)$

$=0.500$ moles HCl

Through assumption, the density, and heat capacity of Dilute solution$=$ That of water in $g/mL$ and $4.184J/g°C$ heat gained by the solution

$\therefore$ $Q= mC_pdT$

$Q=(500.0mL\times 1g/mL)\times (4.184J/g°C )\times(21.8-19.2°C)$

$=0.0561L\times 12mol/L$

$=0.6192molHCl$