Depression in freezing point (Δ

) =

×m×i,
where,

= cryoscopic constant =

,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For

)
Thus, (Δ

) = 1.86 X 0.0085 X 2 =

Now, (Δ

) =

- T
Here, T = freezing point of solution

= freezing point of solvent =

Thus, T =

- (Δ

) = -
<span>Ionic bonding between sodium and phosphate ions.</span>
Impure substance, because pure would be dirt from the earth
<u>Given:</u>
Enthalpy change ΔH for the thermite reaction = -850 kJ
Moles of Al involved = 4
<u>To determine:</u>
Reaction enthalpy when 4 moles of Al reacts
<u>Explanation:</u>
The thermite reaction is-
2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)
When 2 moles of Al react the enthalpy change is -850 kJ
therefore, when 4 moles of Al reacts, the change in enthalpy is-
= 4 moles * (-850) kJ/2 moles = -1700 kJ
Ans: Enthalpy change is -1700 kJ