SATHEE: Chemical Thermodynamics Question 206

Chemical Thermodynamics Question 206

Question: Heat of neutralisation for the given reaction $N a O H + H C l \to N a C l + H_{2} O$ is $57.1, k J, m o l^{- 1}$ . What will be the heat released when $0.25, m o l e$ of $N a O H$ is titrated against $0.25, m o l e$ of $H C l$ [CPMT 1990]

Options:

A) $22.5, k J, m o l^{- 1}$

B) $57.1, k J, m o l^{- 1}$

C) $14.3, k J, m o l^{- 1}$

D) $28.6, k J, m o l^{- 1}$

Show Answer

Answer:

Correct Answer: C

Solution:

$57.1 \times 0.25 = 14.3$ $k J, m o l^{- 1}$ .

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Chemical Thermodynamics Question 206

Question: Heat of neutralisation for the given reaction NaOH+HCl→NaCl+H2O is 57.1,kJ,mol−1 . What will be the heat released when 0.25,mole of NaOH is titrated against 0.25,mole of HCl [CPMT 1990]

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Answer:

Solution:

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Question: Heat of neutralisation for the given reaction $N a O H + H C l \to N a C l + H_{2} O$ is $57.1, k J, m o l^{- 1}$ . What will be the heat released when $0.25, m o l e$ of $N a O H$ is titrated against $0.25, m o l e$ of $H C l$ [CPMT 1990]