Question

A solution is made by dissolving 48.5 g of NaCl in 300.0 g acetic acid. what is the molal concentration of the solution?

Answer 1

mass of acetic acid =300g

density of pure acetic acid is = $1.05g/cm^3$

volume of acetic acid $=mass/density=300/1.05=285.71cm^3=0.28571litres$

moles of NaCl =$48.5/58.45=0.8298moles$

Molar concentration in this case $=0.8298/0.28571=2.904M$