One mole of anhydrous salt A B dissolves in water and liberates 21.0 J mol -1 of heat. The value of Δ H text (hydration) of A B is -29.4 J mol -1 . The heat of dissolution of hydrated salt, A B .2 H 2 O (s) is

Question

One mole of anhydrous salt A B dissolves in water and liberates 21.0 J mol -1 of heat. The value of Δ H text (hydration) of A B is -29.4 J mol -1 . The heat of dissolution of hydrated salt, A B .2 H 2 O (s) is (A) 50.4 J mol-1 (B) 8.4 J mol-1 (C) -50.4 J mol-1 (D) -8.4 J mol-1

Tardigrade Admin · Accepted Answer

Δ H text solution=Δ H text dissolution+Δ H text hydration -21.0=Δ H text dissolution+(-29.4) Δ H text dissolution =-21.0+29.4=8.4 J mol-1

Q. One mole of anhydrous salt $A B$ dissolves in water and liberates $21.0 J m o l^{- 1}$ of heat. The value of $Δ H_{(hydration)}$ of $A B$ is $- 29.4 J m o l^{- 1} .$ The heat of dissolution of hydrated salt, $A B .2 H_{2} O_{(s)}$ is

$50.4 J m o l^{- 1}$

$8.4 J m o l^{- 1}$

$- 50.4 J m o l^{- 1}$

$- 8.4 J m o l^{- 1}$

$Δ H_{solution} = Δ H_{dissolution} + Δ H_{hydration}$ $- 21.0 = Δ H_{dissolution} + (- 29.4)$

$Δ H_{dissolution} = - 21.0 + 29.4 = 8.4 J m o l^{- 1}$

Q. One mole of anhydrous salt AB dissolves in water and liberates 21.0Jmol−1 of heat. The value of ΔH(hydration) ​ of AB is −29.4Jmol−1. The heat of dissolution of hydrated salt, AB.2H2​O(s)​ is

50.4Jmol−1

8.4Jmol−1

−50.4Jmol−1

−8.4Jmol−1