Consider the given data :(a) HCl_{(g)} + 10 | vedclass

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(B) From the given data,the enthalpy change $\Delta H$ is negative,which indicates that the dissolution of $HCl_{(g)}$ in water is an exothermic proce

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The heat of solution depends on the amount of solvent.

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(B) From the given data,the enthalpy change $\Delta H$ is negative,which indicates that the dissolution of $HCl_{(g)}$ in water is an exothermic process.The heat of solution varies depending on the amount of solvent used,as seen by the different $\Delta H$ values for different amounts of water.To find the heat of dilution,we subtract equation $(I)$ from equation $(II)$:$(HCl_{(g)} + 40 \ H_2O_{(l)}) - (HCl_{(g)} + 10 \ H_2O_{(l)}) \rightarrow (HCl \cdot 40 \ H_2O) - (HCl \cdot 10 \ H_2O)$$HCl \cdot 10 \ H_2O + 30 \ H_2O_{(l)} \rightarrow HCl \cdot 40 \ H_2O$$\Delta H_{dilution} = \Delta H_2 - \Delta H_1 = -72.79 \ kJ \ mol^{-1} - (-69.01 \ kJ \ mol^{-1}) = -3.78 \ kJ \ mol^{-1}$.Therefore,option $(B)$ is the correct statement.

Consider the given data $:$
$(a) \ HCl_{(g)} + 10 \ H_2O_{(l)} \rightarrow HCl \cdot 10 \ H_2O \quad \Delta H = -69.01 \ kJ \ mol^{-1}$
$(b) \ HCl_{(g)} + 40 \ H_2O_{(l)} \rightarrow HCl \cdot 40 \ H_2O \quad \Delta H = -72.79 \ kJ \ mol^{-1}$
Choose the correct statement $:$

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Consider the given data $:$$(a) \ HCl_{(g)} + 10 \ H_2O_{(l)} \rightarrow HCl \cdot 10 \ H_2O \quad \Delta H = -69.01 \ kJ \ mol^{-1}$$(b) \ HCl_{(g)} + 40 \ H_2O_{(l)} \rightarrow HCl \cdot 40 \ H_2O \quad \Delta H = -72.79 \ kJ \ mol^{-1}$Choose the correct statement $:$