One mole of anhydrous `MgCl_(2)` dissolves in water and librates `25 cal//mol` of heat. `Delta H_("hydration")` of `MgCl_(2)=30 cal//mol`. Heat of dissolution of `MgCl.H_(2)O`

Options

A. `+5` cal/mol
B. `-5` cal/mol
C. 55 cal/mol
D. `-55` cal/mol

Answer

A

Explanation

`MgCl_(2).H_(2)O rarrMgCl_(2)(aq)Delta H_(1)= ?`
`MgCl_(2)(aq)rarrMg^(2+)(aq)+2Cl^(-)(aq)Delta H_(2)= -30 cal//mol`
`MgCl_(2)(s)rarrMg^(2+)(aq)+2Cl^(-1)(aq)Delta H_(3)= -25 cal//mol`
`Delta H_(1)+Delta H_(2)=Delta H_(3)`
`Delta H_(1)=Delta H_(3)-Delta H_(2)= -25+30= +5 cal//mol`