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(A) Step $1$: Determine the enthalpy of vaporization $(\Delta H_{vap})$ for $1 \ mol$ of water.
Subtract the second equation from the first:
$H_2O_{

Vedclass · Accepted Answer

$5.65$

Vedclass · Answer

(A) Step $1$: Determine the enthalpy of vaporization $(\Delta H_{vap})$ for $1 \ mol$ of water.
Subtract the second equation from the first:
$H_2O_{(l)} \longrightarrow H_2O_{(g)} \quad \Delta H_{vap} = (-57 \ kCal) - (-68.3 \ kCal) = 11.3 \ kCal/mol$.
Step $2$: Calculate the number of moles in $9 \ g$ of water.
Molar mass of $H_2O = 18 \ g/mol$.
$	ext{Moles} = \frac{9 \ g}{18 \ g/mol} = 0.5 \ mol$.
Step $3$: Calculate the total heat required.
$	ext{Heat} = 	ext{Moles} 	imes \Delta H_{vap} = 0.5 \ mol 	imes 11.3 \ kCal/mol = 5.65 \ kCal$.
Therefore,the correct option is $A$.

Calculate the heat required to convert $9 \ g$ of liquid water to water vapor using the following equations:
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(g)} \quad \Delta H = -57 \ kCal$
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(l)} \quad \Delta H = -68.3 \ kCal$ (in $kCal$)

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Calculate the heat required to convert $9 \ g$ of liquid water to water vapor using the following equations: $H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(g)} \quad \Delta H = -57 \ kCal$ $H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(l)} \quad \Delta H = -68.3 \ kCal$ (in $kCal$)