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

Step-by-step explanation:

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

Mathematically, Density =
`(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 =
`(mass)/(volume)`

1.00 g/ml =
`(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 * 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.