After water reaches its boiling point, an additional 2257 J/g isrequired to vapo

Question

After water reaches its boiling point, an additional 2257 J/g isrequired to vaporize it. Determine the mass of water that isvaporized. What volume will it occupy at 100.0 o C whenp=750 torr?

Here are the calculations I made previous to this:
I have 6.05x106g H2O
The temperature, starting out at this problem is174oC
The heat put out by the reaction previous to this problem was1.93x107 kJ = 1.93 x 1010J

Please help, be sure to show the equations, and not just theanswer, because that's not going to help me understand theproblem.

Thanks!

Explanation / Answer

Mass of water = 6.05x106g H2O Heat of vaporization = 2257 J/g Heat given to the water = 1.93 x 1010J Mass of water that vaporizes = 1.93 x 1010J *   ( 1   g  /   2257 J)                                               = 8.55 * 106 g . . Moles of water = mass / molar mass                          = 8.55 * 106 g / 18 g/mol                         = 4.75 *105    mol . P = 750 torr = 0.9868 atm T = 100 C = 373 K . Volume, V = n RT / P                    = 4.75 *105    mol * 0.0821 L-atm/mol/K * 373K / 0.9868 atm                    = 1.47 * 107L

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