If the melting point of ice is 0°C and latent heat of fusion is 5.46 kJ mol-1, the entropy change for the conversion of 18 g of ice to water at one atmosphere is

Question

If the melting point of ice is 0°C and latent heat of fusion is 5.46 kJ mol-1, the entropy change for the conversion of 18 g of ice to water at one atmosphere is (A) 5.46 J mol-1 K-1 (B) 20 J mol-1 K-1 (C) 50 J mol-1 K-1 (D) 360 J mol-1 K-1

Tardigrade Admin · Accepted Answer

Δ Sf =(Δ Hf/Tm.p.) =(5.46×1000/273) =20 J mol-1 K-1

Q. If the melting point of ice is $0^{°} C$ and latent heat of fusion is $5.46 k J m o l^{- 1}$ , the entropy change for the conversion of $18 g$ of ice to water at one atmosphere is

$5.46 J m o l^{- 1} K^{- 1}$

$20 J m o l^{- 1} K^{- 1}$

$50 J m o l^{- 1} K^{- 1}$

$360 J m o l^{- 1} K^{- 1}$

$Δ S_{f} = \frac{Δ H _{f}}{T _{m . p .}}$
$= \frac{5.46 \times 1000}{273}$
$= 20 J m o l^{- 1} K^{- 1}$

Q. If the melting point of ice is 0°C and latent heat of fusion is 5.46kJmol−1, the entropy change for the conversion of 18g of ice to water at one atmosphere is

5.46Jmol−1K−1

20Jmol−1K−1

50Jmol−1K−1

360Jmol−1K−1