Calculate the mass of water sufficient to cover the surface of the Earth with 1" of liquid. (This is the equivalent amount of water vapor contained in the Earth's atmosphere at any instant in time). You need to know that 1" is equivalent to 0.0254 m, the radius (R) of the Earth is 6,370,000 m, the surface area of a sphere is 4πR2, and the density of water is 1000 kg m-3. (You may approximate π as 3.14).