Heat of solution of `BaCl_2.2H_2O = 200 kJ//mol`, Heat of hydration of `BaCl_2 = –150 kJ//mol`. Hence, heat of solution of `BaCl_2` is:

Options

A. 350 kJ
B. 50 kJ
C. `-350 kJ`
D. `-50 kJ`

Explanation

We have
`BaCl_(2). 2H_(2)O+aq rarr Ba^(2+)(aq)+2Cl^(-)(aq)+2H_(2)O , " " Delta H = 200 kJ mol^(-1)`
`BaCl_(2)+2H_(2)O rarr BaCl_(2).2H_(2)O " " Delta H=- 150 kJ mol^(-1)`
On adding both equation we get
`BaCl_(2)+aq rarr Ba^(2+)(aq)+2Cl^(-1)(aq) " " Delta H=50 kJ mol^(-1)`
This is the reaction for dis solution of `BaCl_(2)`.
Therefore heat of this reactions is in fact the heat of solution of `BaCl_(2)`.