40 gm of $N a O H$ reacts with 24.5 gm $H_{2} S O_{4}$ in an aqueous medium. The quantity of heat liberated due to neutralisation is:

13.7 K.cal

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27.4 K.cal

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6.85 K.cal

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1.37 K.cal

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Solution

The correct option is C 6.85 K.cal
Moles of NaOH $= \frac{40}{40} = 1$
Moles of $H_{2} S O_{4} = \frac{24.5}{98} = \frac{1}{4}$
Moles of $O H^{- 1} = 1$
Moles of $H^{+} = 1 / 2$
$∴$ Heat liberated will be $13.7 \times \frac{1}{2}$ Kcal (as $H^{+}$ is limiting) $=$ 6.85 Kcal

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40 gm of NaOH reacts with 24.5 gm H2SO4 in an aqueous medium. The quantity of heat liberated due to neutralisation is:

The correct option is C 6.85 K.calMoles of NaOH =4040=1Moles of H2SO4=24.598=14Moles of OH−1=1Moles of H+=1/2∴ Heat liberated will be 13.7×12 Kcal (as H+ is limiting) = 6.85 Kcal

40 gm of $N a O H$ reacts with 24.5 gm $H_{2} S O_{4}$ in an aqueous medium. The quantity of heat liberated due to neutralisation is:

The correct option is C 6.85 K.cal
Moles of NaOH $= \frac{40}{40} = 1$
Moles of $H_{2} S O_{4} = \frac{24.5}{98} = \frac{1}{4}$
Moles of $O H^{- 1} = 1$
Moles of $H^{+} = 1 / 2$
$∴$ Heat liberated will be $13.7 \times \frac{1}{2}$ Kcal (as $H^{+}$ is limiting) $=$ 6.85 Kcal