Answer:
ΔS(23g NH₃) = 132.2 j/K ≅ 130 j/K (~2 sig. figs. based on 2 sig. figs. in 23 grams NH₃).
Explanation:
Given ΔHv(NH₃) = 23.4 Kj/mol = 23,400 joules/mole
T = -33.5°C = (-33.5 + 273)K = 239.5K
molar ΔS (at boiling) = ΔHv/T = 23,400 j/mol / 239.5K = 97.70 j/mol·K
ΔS(23g NH₃) = molar ΔS (at boiling) x (moles NH₃) = 97.70j/mol·K x (23g/17g·mol⁻¹) = 132.2 j/K ≅ 130 j/K (~2 sig. figs. based on 2 sig. figs. in 23 grams NH₃).