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Relative Lowering of Vapour Pressure

The enthalpy ...

Question

The enthalpy change on freezing $1$ mol of water at $5^{o} C$ to ice at $- 5^{o} C$ is:
(Given: $Δ_{f u s} H = 6 k J {m o l}^{- 1}$ at $0^{o} C$
$C_{P} (H_{2} O, l) = 75.3 J {m o l}^{- 1}$ $K^{- 1}$ ,
$C_{P} (H_{2} O, s) = 36.8 J {m o l}^{- 1}$ $K^{- 1}$ )

- 6.56 k J {m o l}^{- 1}

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- 5.87 k J {m o l}^{- 1}

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6.0 k J {m o l}^{- 1}

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5.44 k J {m o l}^{- 1}

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Solution

The correct option is D $- 5.87 k J {m o l}^{- 1}$
$△ H = C p △ T$ (for changing from $5 - 0 ° C$ )
$= 75.3 \times 5 = 376.5 J = 0.377 K J$
$△ H = C p △ T$ (for changing from $0 - 5 ° C$ )
$= 36.8 \times 5 = - 184 J = - 0.184 K J$
Total enthalpy change $= 0.377 + (- 6.0) + (- 0.184) = - 5.807 K J / m o l$

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Similar questions

1 m o l

5^{o} C

- 5^{o} C

(

Δ_{f u s} H = 6 k J m o l^{- 1}

0^{o} C

C_{p} (H_{2} O, l) = 75.3 J m o l^{- 1} K^{- 1}

C_{p} (H_{2} O, S) = 36.8 J m o l^{- 1} K^{- 1})

5^{o} C

- 5^{o} C

[{G i v e n : △}_{f u s} H = 6 k J {m o l}^{- 1} a t 0^{o} C, C_{p} (H_{2} O, l) = 75.3 J {m o l}^{- 1} K^{- 1}, C_{p} (H_{2} O, s)

= 36.8 J {m o l}^{- 1} K^{- 1}]

5^{0} C

- 5^{0} C

(Given $Δ_{f u s} H = 6 k J m o l^{- 1} a t 0^{0} C$ ,

$C_{p} (H_{2} O, 1) = 75.3 J m o l^{- 1} K^{- 1}$

$C_{p} (H_{2} O, s) = 36.8 J m o l^{- 1} K^{- 1})$ )

1

5^{o} C

- 5^{o} C

Δ_{f u s} H = 6.03 k J m o l^{- 1}

0^{o} C

C_{p} [H_{2} O (l)] = 75.3 J m o l^{- 1} K^{- 1}

C_{p} [H_{2} O (s)] = 36.8 J m o l^{- 1} K^{- 1}

1.0

{- 10.0}^{o} C, Δ_{f u s} H = 6.03 k J {m o l}^{- 1}

0^{o} C

C_{P} [H_{2} O (l)] = 75.3 J {m o l}^{- 1} K^{- 1}

C_{P} [H_{2} O (s)] = 36.8 J {m o l}^{- 1} K^{- 1}

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