Question

A 70.0 mL sample of water is heated to its boiling point. How much heat is required to vaporize it? (Assume a density of 1.00 g/mL.)

Answer 1

158 kJ. if you convert into moles and then divide by the number on the table you should get this

Answer 2

Explanation:

As density is the amount of mass divided by volume of the substance.

Mathematically,     Density = $\frac{mass}{volume}$

It is given that volume is 70.0 ml and density is 1.00 g/ml. Therefore, mass of the given substance will be as follows.

                Density = $\frac{mass}{volume}$

                 1.00 g/ml = $\frac{mass}{70.0 ml}$

                        mass = 70.0 g

As we known that heat of vaporization of water is 2260 J/g for 1 g of a substance. Therefore, heat of vaporization of water for 70.0 g will be as follows.

                     $70.0 g \times 2260 J/g$

                       = 158200 J

or,                     = 158.2 kJ                (as 1 kJ = 1000 J)

Thus, we can conclude that 158.2 kJ heat is required to vaporize given sample of water.