The enthalpy change on freezing of 1 mole of water at 5°C to ice at -5°C in kJ mol^-1 is
Given, ${ΔH}_{fusion} = 6 KJ {mol}^{- 1} at 0^{0} C$
$C_{P} (H_{2} O, l) = 75 .3 J {mol}^{- 1} K^{- 1}$
$C_{P} (H_{2} O, S) = 36 .8 J {mol}^{- 1} K^{- 1}$

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Detailed Solution

$\underset{(H_{2} O, l)}{5^{0} C} \overset{{ΔH}_{1}}{\to} \underset{(H_{2} O, l)}{0^{0} C} \overset{{ΔH}_{2}}{\to} \underset{(H_{2} O, S)}{0^{0} C} \overset{{ΔH}_{3}}{\to} \underset{(H_{2} O, S)}{- 5^{0} C}$

$ΔH = {nC}_{p} ΔT$

${ΔH}_{2} = enthalpy of fusion$

$ΔH = {ΔH}_{1} + {ΔH}_{2} + {ΔH}_{3}$

$Δ H = {n_{C}}_{p} d T + Δ H_{f} + n_{C_{p} d T}$

$ΔH = (1 \times 75 .3 \times 5) + (6000) + (1 \times 36 .8 \times 5) J$

$= 6 .560 KJ / mol / K$

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