The dissolution of 1 mole of NaOH(s) in $100$ mole of $H_{2} O (l)$ give rise to evolution of heat as -42.34 kJ. However, if 1 mole of NaOH(s) is dissolved in $1000$ mole of $H_{2} O (l)$ the heat given out is 42.76 kJ. If the enthalpy change when $900$ mole of $H_{2} O (l)$ are added to a solution containing 1 mole of NaOH(s) in $100$ mole of $H_{2} O$ is $x$ $k J$ then the value of $- 100 x$ is _____.

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Solution

$N a O H (s) + 100 H_{2} O ⟶ N a O H (100 H_{2} O); Δ H = - 42.34 k J$ ... (i)
$N a O H (s) + 1000 H_{2} O ⟶ N a O H (1000 H_{2} O); Δ H = - 42.76 k J$ ... (ii)
By subtracting equations $[(i i) - (i)]$ , we get-
$N a O H (s) + 900 H_{2} O ⟶ N a O H (900 H_{2} O); Δ H = - 0.42 k J$

$∴ - 100 x = 42$

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NaOH(s)+100H2O⟶NaOH (100H2O);ΔH=−42.34kJ ... (i)NaOH(s)+1000H2O⟶NaOH (1000H2O);ΔH=−42.76kJ ... (ii)By subtracting equations [(ii)−(i)], we get-NaOH(s)+900H2O⟶NaOH (900H2O);ΔH=−0.42kJ∴−100x=42