Question

The heat of vaporization for water is 40. 7 kJ/mol. How much heat energy must 150. 0 g of water absorb to boil away completely? Use the formula mc001-1. Jpg. 339. 2 kJ 381. 6 kJ 610. 5 kJ 6,105 kJ.

Answer 1

The heat absorbed by the water sample for vaporization has been 6,105 kJ. Thus, option D is correct.

The heat of vaporization has been the amount of heat required to vaporize 1 gram of liquid.

The heat required for the vaporization has been given as:

$Q=m\Delta H_{vap}$

Computation for the Heat of vaporization

The heat of vaporization of water has been given as, $\Delta H_{vap}=40.7\;\rm kJ/mol$

The mass of water sample has been, $m=150\;\rm g$

Substituting the values for the heat energy absorbed, Q:

$Q=150\;\times\;40.7\;\rm kJ\\ \textit Q=6,105\;kJ$

The heat absorbed by the water sample for vaporization has been 6,105 kJ. Thus, option D is correct.

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