Question

Use the table on the right to calculate each required quantity.

The quantity of heat required to melt 175.0 g Cu: ____ kJ

The mass of a sample of NaCl that requires 450.1 kJ of heat to melt fully: ____ g

The identity of a substance that releases 21.2 kJ of energy when 1.42 mol freezes: ____

Answer 1

a. 35.8 KJ

b. 871 g

c. Fe

..........................................................................................

Answer 2

• The amount of heat required to melt 175 g Cu has been 35.8 kJ/mol.

• The mass of NaCl melted with 450.1 kJ heat has been 870.98 grams.

• The substance that has been required 14.9 kJ of energy for freezing 1 mole of sample has been Fe.

The energy required to melt 1 mole of a substance has been given as the heat of fusion. The heat energy required to vaporize 1 mole of a substance has been heat of vaporization.

Heat of vaporization

• The amount of heat required to melt 1 mole of Cu has been 13 kJ/mol. The moles of Cu equivalent to 175 gram has been:

$\rm Moles=\dfrac{Mass}{Molar\;mass}\\Moles \;Cu =\dfrac{175}{63.5}\\Moles\;Cu=2.755\;mol$

The available moles of Cu has been 2.755 mol.

The heat for melting 1 mole of Cu has been 13 kJ/mol. The heat for melting 2.755 mole of Cu has been:

$\rm 1\;mol=13\;kJmol\\2.755\;mol=2.755\;\times\;13\;kJ/mol\\2.755\;mol= 35.8\;kJ/mol$

The amount of heat required to melt 175 g Cu has been 35.8 kJ/mol.

• The amount of heat that can melt 1 mole of NaCl has been 30.2 kJ/mol. The amount of sample melted by 450.1 kJ/mol heat has been:

$\rm 30.2\;kJ/mol=1\;mol\;NaCl\\450.1\;kJ/mol=\dfrac{1}{30.2}\;\times\;450.1\;mol\;NaCl\\450.1\;kJ/mol=14.90\;mol\;NaCl$

The moles of NaCl sample melted has been 14.90 mol. The mass of NaCl sample has been:

$\rm Mass=Moles\;\times\;Molar\;mass\\Mass\;NaCl=14.9\;\times\;58.44\\Mass\;NaCl=870.98\g$

The mass of NaCl melted with 450.1 kJ heat has been 870.98 grams.

• The energy required by an atom to freeze a 1.42 mol sample has been 21.2 kJ/mol. The energy of 1 mole of sample has been:

$\rm 1.42\;mol=21.2\;kJ\\1\;mol=\dfrac{21.2}{1.42} \\1\;mol=14.92\;kJ$

The substance that has been required 14.9 kJ of energy for freezing 1 mole of sample has been Fe.

Learn more about heat of fusion, here:

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