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When \mathrm{600 \mathrm{~mL}} of \mathrm{0.2\; \mathrm{M} \; \mathrm{HNO}_{3}} is mixed with \mathrm{400 \mathrm{~mL}} of \mathrm{0.1 \; \mathrm{M} \; \mathrm{NaOH}}  solution in a flask, the rise in temperature of the flask is ____________\mathrm{\times 10^{-2}{ }^{\circ} \mathrm{C}}.

\mathrm{\text { (Enthalpy of neutralisation }=57 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { and Specific heat of water }=4.2 \mathrm{JK}^{-1} \mathrm{~g}^{-1} \text { ) }}

 (Neglect heat capacity of flask) 

Option: 1

54


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The reaction will be

           \mathrm{HNO_{3}+NaOH\longrightarrow H_{2}O}
           \mathrm{0.2M}            \mathrm{0.1M}

initial  120 m mol         40 m mol        0
final    80  m mol            0                  40 m mol

So, total 40 m mol is neutralising.
So total enthalpy of neutralisation for 40 m mol will be

\mathrm{\Delta H= 40\, m\, mol\times 57\, KJ\, mol}
\mathrm{\Delta H= 40\times 10^{-3}\times 57\times 10^{3}\, J}
\mathrm{\Delta H= 2280\, J}

we know the formula,
\mathrm{Q= ms\Delta T= \Delta H}
m = mass of water
s = specific heat of water  = \mathrm{4.2\, J\, K^{-1}\, g^{-1}}
\mathrm{\Delta T= ?}

\mathrm{Now, m= volume\times density= 1000\, mL\times 1\, g|ml}
            \mathrm{m= 1000\, g}

\mathrm{then, 2280= 1000\times 4.2\times\Delta T }
              \mathrm{\Delta T= 0.54 } 
              \mathrm{\Delta T= 54\times 10^{-2} K}
              \mathrm{\Delta T= 54\times 10^{-2} }\: ^{\circ}C\mathrm{\left ( \text{change will be same }\right )}

Ans = 54



 

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Gunjita

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