Question

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

पर्याय

• 1059 K

• 30.5 K

• 28.8 K

• 28800 K

Answer 1

One mole of NaCl_(s) on melting absorbed 30.5 kJ of heat and its entropy increased by 28.8 J K⁻¹. The melting point of NaCl is 1059 K.

Explanation:

\[\ce{NaCl_{(s)} ⇌ NaCl_{(l)}}\]

Given that: ∆H = 30.5 kJ mol⁻¹

∆S = 28.8 J K⁻¹ = 28.8 × 10⁻³ kJ K⁻¹

By using ∆S = `(∆"H")/"T"`

∴ T = `(30.5  "kJ mol"^-1)/(28.8 xx 10^-3  "kJ K"^-1)` = 1059 K