JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
\(17.0\, g\) of \(NH _{3}\) completely vapourises at \(-33.42^{\circ} C\) and \(1\, bar\) pressure and the enthalpy change in the process is \(23.4 \,k\,J\, mol ^{-1}\). The enthalpy change for the vapourisation of \(85 g\) of \(NH _{3}\) under the same conditions is \(.....\, kJ .\)
- A \(81\)
- B \(117\)
- C \(453\)
- D \(751\)
Answer & Solution
Correct Answer
(B) \(117\)
Step-by-step Solution
Detailed explanation
Given data is for \(1 \,moles\) and asked for \(5\, moles\) so value is \(23.4 \times 5=117\, kJ\)
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