If you could answer 8,9,10,11ab that would be great! Thank you! Consider the fol
ID: 488830 • Letter: I
Question
If you could answer 8,9,10,11ab that would be great! Thank you! Consider the following dication (Chemical formula: C_13H_9N_3ClPd). If you had ~0.300 g total and you wanted to know the concentration of metal in a known solution (perhaps to verity your projected structure), you decked to use Flame-AA. Describe (in sufficient detail that I could walk Into a lab with an AA with a vent stack and do it) how to prepare a solution of 10 - 30 ppm aqueous pd^2+ by mass. You may treat the solution to have a density identical to water The AA has a ppm resolution to the tenths place, so make sure everything else you use has a precision better than that. You may use a balance, any common reagent 1, 2, 5, 10, 15, 25 mL volumetric pipets and 25, 50, 100, 200, 250 mL volumetric flasks. If a sample of baby food, 253 g (with carrots grown near Los Alamos), is found to contain 1.27 ppm thorium ions, how much thorium will bady ingest if she eats the whole container? Running salts (ion pairs) on normal silica plates is pretty useless. Why? When obtaining a spectrum using "proton NMR": One must use duterated solvents. Why? If no Deuterated solvents are available, propose a solvent that one could use (yes, it must be liquid).Explanation / Answer
8)preparation of 10-30ppm pb2+ solution
concentration of Pb2+ to be prepared=10-30ppm=10-30 mg/L=(10-30) *10^-3 g/L=[(10-30) *10^-3 g/L]/molar mass of Pb2+=[(10-30) *10^-3]g/L / 106.42 g/mol=(0.0940-0.282) *10^-3]mol/L
As 1 mole of the given compound has 1 mole of Pb2+,so same moles of the compound is to be dry weighed and dissolved in 1L water.
the compound formula-C13H9N3ClPd
molar mass=13*12.01 g/mol+9*1.008g/mol+3*14.007 g/mol+35.453g/mol+106.42 g/mol=156.13+9.072+42.021+35.453+106.42=349.096 g/mol
So mass of the compound to be dissolved=(0.0940-0.282) *10^-3]mol/L*molar mass of the compound=(0.0940-0.282) *10^-6]mol/L*349.096 g/mol=(32.815-98.445)*10^-3] g/L=0.0328-0.0984 g/L
As it is difficult to weigh so less mass using a balance so higher molarity solution may be prepared and then diluted to the required molarity.
let M1=(0.0940-0.282) *10^-3]mol/L for the given compound [target molarity]
V1=250 ml (say)
M2 =stock solution
V2=10 ml
M1*V1=M2*V2
so,M2=M1*V1/V2=(0.0940-0.282) *10^-3]mol/L *250ml/10 ml=(0.0940-0.282) *25 *10^-3]mol/L =(2.35-7.05)*10^-3 mol/L=
=(2.35-7.05)*10^-3 mol/L*349.096 g/mol=(820.375-2461.127)*10^-3 g/L=0.820-2.46 g/L
M2=0.820-2.46 g/L (stock to be prepared) ,volume =10ml
10 ml of this solution to be pipetted in a 250 ml volumetric flask and diluted upto the mark to prepare the target solution of 10-30 ppm .