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

Answer:

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.

Explanation:

Mass of benzene = 7.35 g

Moles of benzene = $\frac{7.35 g}{78 g/mol}=0.09423 mol$

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

$Q=\Delta H_{fus}\times 0.09423 mol$

$=9.92 kJ/mol\times 0.09423 mol=0.9348 kJ=934.8 J$

(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

$Q=\Delta H_{Vap}\times 0.09423 mol$

$=30.7 kJ/mol\times 0.09423 mol=2.893 kJ=2,993 J$

(1 kJ = 1000 J)