Chemical Thermodynamics Question 206

Question: Heat of neutralisation for the given reaction $ NaOH+HCl\to NaCl+H_2O $ is $ 57.1,kJ,mo{{l}^{-1}} $ . What will be the heat released when $ 0.25,mole $ of $ NaOH $ is titrated against $ 0.25,mole $ of $ HCl $ [CPMT 1990]

Options:

A) $ 22.5,kJ,mo{{l}^{-1}} $

B) $ 57.1,kJ,mo{{l}^{-1}} $

C) $ 14.3,kJ,mo{{l}^{-1}} $

D) $ 28.6,kJ,mo{{l}^{-1}} $

Show Answer

Answer:

Correct Answer: C

Solution:

$ 57.1\times 0.25=14.3 $ $ kJ,mo{{l}^{-1}} $ .



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक