If $∆{\mathrm{V}}_{\mathrm{mix}}<0, \mathrm{then} \mathrm{it} \mathrm{shown}$

If  $∆{\mathrm{V}}_{\mathrm{mix}}<0, \mathrm{then} \mathrm{it} \mathrm{shown}$ negative deviation from Raoult’s law. The reason is the interaction forces between A and B are greater than A-A and B-B. So, during the dissolution process volume is decreased, $∆{\mathrm{V}}_{\mathrm{mix}}<0$