The surface of a solid sphere is covered by a monolayer of receptors for a ligand. When the ligands diffuse to the surface of the sphere, they are captured instantaneously by the receptors. The ligand diffusion coefficient is 5.2*10-9m2/s. Assume the bulk concentration far away from the sphere is constant of 6.6 nM and the sphere radius is 9.8 μm. (a) Find the ligand concentration C at the location of r=4.9 μm at steady state, where r is the distance to the surface of the sphere. Please enter the numerical value with a unit of nM.

The ligand concentration C is 0.274 nM

Explanation:

According to the image, the bulk concentration is inversely proportional to the radius, thus:

Where

Bα = 6.6 nM

r = 14.7 um = 1.47x10⁻⁵m

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1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%