Grains of fine California beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide, which has a density of 2.6 × 103 kg/m3. What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 1.1 m on an edge?
2. The attempt at a solution
1) First, find the surface area of the cube:
A = 6s² = 6 * (1.1 m)² = 7.26 m²
2) Next, find the surface area of one sand grain:
A = 4∏r² = 4∏ * (50 µm)² = 31,416 µm²
3) How many grains of sand equal the area of the cube.
7.26 m²/ 31,416 µm² * (1,000,000 µm/m)² = 231,092,437 grains of sand
4) Find the volume of one grain of sand.
V = (4/3)∏r³ = (4/3)∏(50 µm)³ = 523,599 µm³
5) How many grains of sand is in that cube weigh 2600 kg.
1.00 m³/523,598 µm³ * (1,000,000 µm/m)³ = 1,909,859,228,000 pieces of sand weighs 2600 kg
6) How much does each piece of sand weigh.
2600 kg/1,909,859,228,000 pieces * 1000 g/kg = 1.36E-6 g
7) How much does 231,092,437 grains of sand weigh?
1.81 E -6 g * 231,092,437 = 315 g
Okay, now the problem is that it says my answer is wrong!? I must have done something wrong. I'm thinking that maybe in step 5 where I rounded the last 1,909,859,228,"000" three digits off. Is this that big of a problem to throw my answer off? If not can you help me find what I did wrong?