What is the enthalpy change when 13.8 grams of H2O changes from a liquid to a solid at zero degrees Celsius? ΔHcondensation = −40.65 kJ/mol ΔHfreezing = −6.03 kJ/mol −4.62 kJ −83.2 kJ −7.87 kJ −31.3 kJ

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What is the enthalpy change when 13.8 grams of H2O changes from a liquid to a solid at zero degrees Celsius? ΔHcondensation = −40.65 kJ/mol ΔHfreezing = −6.03 kJ/mol −4.62 kJ −83.2 kJ −7.87 kJ −31.3 kJ

Chemistry
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how many moels of water is in 13.8g of H2O?
there is 1.22 moles
no. 1mol of H2O is 18g. You've only got 13.8g. Flip the fraction upside down

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okay so the answer would −7.87 kJ ?
or am i wrong?
your mole conversion should look like\[13.8g H_2O * (\frac{1mol H_2O}{18gH_2O}) = 0.767 moles H_2O\]
so is my answer wrong?
mole of water "costs" 40.65kJ of energy to freeze. You've got only 76.7% of 1 mole. The heat of fusion for .767mol of H2O will be \[0.767mol H_2O * (\frac{40.65kJ}{1mol H_2O}) = 31.17kJ\]
40.65 or 6.03 ????
Oh okay . −31.3 kJ so this would be the final result?
@mos1635 is correct, replace the 40.65kJ with 6.03kJ. I missed the condensation vs freezing.
-4.623\[\approx\] -4.62

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