When 600 ; mL of 0.2 ; M ; HNO_3 is mixed wit | vedclass

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(B) The reaction is: $HNO_3 + NaOH ightarrow NaNO_3 + H_2O$Initial moles of $HNO_3 = 600 \; mL 	imes 0.2 \; M = 120 \; m \; mol = 0.12 \; mol$Initi

Vedclass · Accepted Answer

$54$

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(B) The reaction is: $HNO_3 + NaOH ightarrow NaNO_3 + H_2O$Initial moles of $HNO_3 = 600 \; mL 	imes 0.2 \; M = 120 \; m \; mol = 0.12 \; mol$Initial moles of $NaOH = 400 \; mL 	imes 0.1 \; M = 40 \; m \; mol = 0.04 \; mol$Since $NaOH$ is the limiting reagent,the moles of water formed $= 0.04 \; mol$.Heat released $(q)$ $= 	ext{moles of water} 	imes \Delta_{neut}H = 0.04 \; mol 	imes 57 	imes 10^3 \; J \; mol^{-1} = 2280 \; J$.Total volume of solution $= 600 \; mL + 400 \; mL = 1000 \; mL$.Assuming density of solution $\approx 1 \; g \; mL^{-1}$,mass of solution $(m)$ $= 1000 \; g$.Using $q = m 	imes S 	imes \Delta T$:$2280 \; J = 1000 \; g 	imes 4.2 \; J \; K^{-1} \; g^{-1} 	imes \Delta T$$\Delta T = \frac{2280}{4200} \; K = 0.54286 \; K = 0.54286 \; ^{\circ}C$Expressing in terms of $10^{-2} \; ^{\circ}C$:$\Delta T = 54.286 	imes 10^{-2} \; ^{\circ}C \approx 54 	imes 10^{-2} \; ^{\circ}C$.

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