JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
When \(600\; mL\) of \(0.2 \;M \;HNO _{3}\) is mixed with \(400\; mL\) of \(0.1\; M\; NaOH\) solution in a flask, the rise in temperature of the flask is \(\dots\;\times 10^{-2}{ }^{0} C\) (Enthalpy of neutralisation \(=57 \;kJ \;mol ^{-1}\) and Specific heat of water \(=4.2\; JK ^{-1} g ^{-1}\) ) (Neglect heat capacity of flask)
- A \(53\)
- B \(54\)
- C \(55\)
- D \(56\)
Answer & Solution
Correct Answer
(B) \(54\)
Step-by-step Solution
Detailed explanation
\(HNO _{3} \quad NaOH\) \(600 \;mL \times 0.2\; M \quad 400 \;mL \times 0.1 \;M\) \(=120\; m \;mol \quad=40 \;m\; mol\) \(HNO _{3}+ NaOH \rightarrow NaNO _{3}+ H _{2} O\) \(\begin{array}{lccc}\text { Bef. } & 120 & 40 \\ \text { Aft. } & 80 & 0 & 40\; m\; mol\end{array}\)…
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