Question

You are a studen in Organic Chemsitry lab and you are tasked with the job to make 202 ppm solution of Rhodamine B (MM = 479.02 g/mol). How many moles of dye do you need to add to a 250 mL beaker of water to prepare this solution?

Answer 1

Given concentration of Rhodamine B is 202 ppm

                                                       = 202 mg/L                     Since 1ppm = 1 mg/L

So the mass of Rhodamine B present in 1 L=1000 mL of the solution is 202 mg

   the mass of Rhodamine B present in 250 mL of the solution is M mg                                   

    M = ( 250x202) / 1000

       = 50.5 mg

       = 50.5x10^-3 g

Therefore the mass of Rhodamine B present in 250 mL is 50.5x10^-3 g

Given the molar mass of Rhodamine B is 479.02 g/mol

So the number of moles of Rhodamine B , n = mass/molar mass

                                                                = (50.5x10^-3 g ) / 479.02 (g/mol)

                                                                = 1.054x10^-4 mol

Therefore 1.054x10^-4 mol of dye needed to add to a 250 mL beaker of water to prepare this solution