#### Answer

369 g

#### Work Step by Step

The amount of heat required to vaporize one mole of liquid is defined as a heat of vaporization. Heat of vaporization is represented by dH vap.
The heat of vaporization of 1 mol of water is 40.7 kJ
835 kJ= 835 kJ x(1 mol/ 40.7 kJ)=20.5 mol
So, with 835 kJ of dH vap we can vaporize 20.5 mol of water.
Since 1mol of water = 18g.
20.5mol of water= 20.5 mol x 18 g= 369 g.
So with 835 kJ of dH, we can vaporize 369 g of water.