17.0 g of $\mathrm{NH}_{3}$ completely vapourises at $-33.42^{\circ}\mathrm{C}$ and 1 bar pressure and the enthalpy change in the process is $23.4\,\mathrm{kJ\,mol}^{-1}$. The enthalpy change for the vapourisation of 85 g of $\mathrm{NH}_{3}$ under the same conditions is ______ kJ.

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Some Basic Concepts of Chemistry

17.0 g of

NH_{3}

completely vapourises at

- 33.4 2^{\circ} C

and 1 bar pressure and the enthalpy change in the process is

23.4 kJ mol^{- 1}

. The enthalpy change for the vapourisation of 85 g of

NH_{3}

under the same conditions is ______ kJ.

Answer: 117

Molar mass of

NH_{3} = 17 g/mol

. So 17.0 g corresponds to 1 mole.

Given:

Δ H_{vap} = 23.4 kJ/mol

for 1 mole.

For 85 g: number of moles

= \frac{85}{17} = 5

moles.

Total enthalpy change

= 5 \times 23.4 = 117 kJ

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer