Question

Calculate the heat required to melt 7.35 g of benzene at its normal melting point. Heat of fusion (benzene) = 9.92 kJ/mol Heat = kJ

Calculate the heat required to vaporize 7.35 g of benzene at its normal boiling point. Heat of vaporization (benzene) = 30.7 kJ/mol Heat = kJ

Answer 1

To melt 7.35 g of benzene, 0.933 kJ of heat is required; to vaporize the same amount, 2.889 kJ of heat is needed. These calculations involve converting the mass of benzene to moles and multiplying by the respective heat of fusion and vaporization values.

Step-by-step explanation:

To calculate the heat required to melt 7.35 g of benzene at its melting point, we use the formula:

Q = n × ΔHfus, where Q is the heat absorbed, n is the number of moles, and ΔHfus is the heat of fusion. The molar mass of benzene (C6H6) is 78.11 g/mol.

First, convert the mass of benzene to moles:

n = mass (g) ÷ molar mass (g/mol) = 7.35 g ÷ 78.11 g/mol = 0.0941 moles

Now, multiply the moles by the heat of fusion:

Q = 0.0941 moles × 9.92 kJ/mol = 0.933 kJ

The heat required to melt 7.35 g of benzene is 0.933 kJ.

To calculate the heat required to vaporize 7.35 g of benzene at its boiling point, we use a similar approach:

n = 7.35 g ÷ 78.11 g/mol = 0.0941 moles (as calculated above)

Q = n × ΔHvap, where ΔHvap is the heat of vaporization.

Q = 0.0941 moles × 30.7 kJ/mol = 2.889 kJ

The heat required to vaporize 7.35 g of benzene is 2.889 kJ.

Answer 2

The heat required to melt 7.35 g of benzene at its normal melting point is 934.8 Joules.

The heat required to vaporize 7.35 g of benzene at its normal melting point is 2,893 Joules.

Step-by-step explanation:

Mass of benzene = 7.35 g

Moles of benzene =
`(7.35 g)/(78 g/mol)=0.09423 mol`

Heat fusion of benzene,
`\Delta H_(fus) = 9.92 kJ/mol`

1) Heat required to melt 7.35 g of benzene at its normal melting point = Q

`Q=\Delta H_(fus)* 0.09423 mol`

`=9.92 kJ/mol* 0.09423 mol=0.9348 kJ=934.8 J`

(1 kJ = 1000 J)

2) Heat vaporization of benzene,
`\Delta H_(vap) = 30.7 kJ/mol`

Heat required to vaporize 7.35 g of benzene at its normal melting point = Q

`Q=\Delta H_(Vap)* 0.09423 mol`

`=30.7 kJ/mol* 0.09423 mol=2.893 kJ=2,993 J`

(1 kJ = 1000 J)