One mole of NaCl (s) on melting absorbed 30.5 kJ of heat and its entropy is increased by 28.8 J K-1. The melting point of NaCl is:

Question

One mole of NaCl (s) on melting absorbed 30.5 kJ of heat and its entropy is increased by 28.8 J K-1. The melting point of NaCl is: (A) 1059 K (B) 30.5 K (C) 28.8 K (D) 28800 K

Tardigrade Admin · Accepted Answer

Use the following formula to solve the problem.

Δ S = \frac{Δ H}{T}

Given,

Δ S =

change in entropy

= 28.8 J K^{- 1}

= 28.8 \times 1 0^{- 3} k J K^{- 1}

Δ H =

change in enthalpy

= 30.5 k J

mol

^{- 1}

r =

temperature

∴ 28.8 \times 1 0^{- 3} = \frac{30.5}{T}

or

T = \frac{30.5}{28.8 \times 1 0 ^{- 3}}

= 1.059 \times 1 0^{3} = 1059 K

Q. One mole of NaCl (s) on melting absorbed 30.5kJ of heat and its entropy is increased by 28.8JK−1. The melting point of NaCl is:

1059 K

30.5 K

28.8 K

28800 K

Use the following formula to solve the problem. ΔS=TΔH​ Given, ΔS= change in entropy =28.8JK−1 =28.8×10−3kJK−1 ΔH= change in enthalpy =30.5kJ mol −1 r= temperature ∴28.8×10−3=T30.5​ or T=28.8×10−330.5​ =1.059×103=1059K

Q. One mole of $N a Cl$ (s) on melting absorbed $30.5 k J$ of heat and its entropy is increased by $28.8 J K^{- 1}$ . The melting point of $N a Cl$ is: