Enthalpy change when 1 g water is frozen at 0° C is: (Δ H text fus =1.435 kcal mol -1)

Question

Enthalpy change when 1 g water is frozen at 0° C is: (Δ H text fus =1.435 kcal mol -1) (A) 0.0797 kcal (B) - 0.0797 kcal (C) 1.435 kcal (D) -1.435 kcal

Tardigrade Admin · Accepted Answer

Enthalpy change per gram water is calculated by dividing Δ H by molecular mass of water. Given Δ H =1.4354 kcal mol -1 =-(1.4354/18)=-0.0797 kcal g -1

Q. Enthalpy change when $1 g$ water is frozen at $0^{\circ} C$ is : $(Δ H_{fus} = 1.435 k c a l m o l^{- 1})$

0.0797 kcal

- 0.0797 kcal

1.435 kcal

-1.435 kcal

Enthalpy change per gram water is calculated by dividing $Δ H$ by molecular mass of water.
Given $Δ H = 1.4354 k c a l m o l^{- 1}$
$= - \frac{1.4354}{18} = - 0.0797 k c a l g^{- 1}$

Q. Enthalpy change when 1g water is frozen at 0∘C is : (ΔHfus ​=1.435kcalmol−1)