Sure, let's walk through the detailed, step-by-step solution for calculating the heat involved in condensing 18.5 g of [tex]\( CH_3OH \)[/tex] (methanol) with a given enthalpy of vaporization, [tex]\(\Delta H_{vap} = 38.0 \, \text{kJ/mol}\)[/tex].
### Step 1: Determine the Molar Mass of Methanol
First, we need the molar mass of methanol [tex]\( CH_3OH \)[/tex]:
[tex]\[ \text{C: } 12.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{H: } 4 \times 1.01 \, \text{g/mol} = 4.04 \, \text{g/mol} \][/tex]
[tex]\[ \text{O: } 16.00 \, \text{g/mol} \][/tex]
Add these together to get the molar mass:
[tex]\[ 12.01 + 4.04 + 16.00 = 32.05 \, \text{g/mol} \][/tex]
### Step 2: Calculate the Number of Moles
Next, we calculate the number of moles of [tex]\( CH_3OH \)[/tex] in 18.5 grams:
[tex]\[ \text{mass of methanol} = 18.5 \, \text{g} \][/tex]
[tex]\[ \text{molar mass of methanol} = 32.05 \, \text{g/mol} \][/tex]
[tex]\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{18.5 \, \text{g}}{32.05 \, \text{g/mol}} \approx 0.577 \, \text{moles} \][/tex]
### Step 3: Calculate the Heat Involved in Condensation
Finally, we use the number of moles we calculated to determine the heat (in kJ) involved in condensing the methanol, given the enthalpy of vaporization ([tex]\(\Delta H_{vap} = 38.0 \, \text{kJ/mol}\)[/tex]):
[tex]\[ \text{heat} = \text{number of moles} \times \Delta H_{vap} \][/tex]
[tex]\[ \text{heat} \approx 0.577 \, \text{moles} \times 38.0 \, \text{kJ/mol} = 21.94 \, \text{kJ} \][/tex]
Therefore, the heat involved in condensing 18.5 g of [tex]\( CH_3OH \)[/tex] is approximately 21.94 kJ.
