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