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MUST SOLVE ALL PARTS FOR FIVE STARS A standard solution for Fe 2+ determination

ID: 789746 • Letter: M

Question

MUST SOLVE ALL PARTS FOR FIVE STARS


A standard solution for Fe2+ determination can be prepared from ferrous ammonium sulfate (Fe(NH4)2(SO4)2.6H2O).

Calculate the formula weight of ferrous ammonium sulfate.  (Units = g/mol)


Calculate the mass of Fe in 151 mg of ferrous ammonium sulfate.


Calculate the mass of ferrous ammonium sulfate you would need to add to a 2.00 mL volumetric flask to make a solution that is 39 ppm in Fe.


Stock solution A is prepared by dissolving 172 mg of ferrous ammonium sulfate in water and diluting to volume in a 500-mL volumetric flask.  Calculate the concentration of iron, in ppm, in stock solution A.


Stock solution B is prepared by diluting 13.00 mL of stock solution A to 100.00 mL. Calculate the concentration of iron, in ppm, in stock solution B.

Explanation / Answer

formula mass = (56 + (14+4*1)*2) + (32+16*4)*2 + 6*(1*2+16) = 392 g/mol.................ans 1


so in 392 gm of ferrous ammonium sulfate contain 56 gm Fe

so in 151mg = 21.57 mg Fe...........................................................ans 2


39ppm = 39mg Fe/ L of solution

39ppm = 39mg/1000ml

so for 2 ml 39 ppm is 0.039*2 = 0.078mg Fe

so 1 moles of Fe is present in 1 mole of ferrous ammonium sulfate

so 0.078mg = 0.078/56 = 0.00139 milimoles Fe

so mili moles of ferrous ammonium sulfate = 0.00139

so mass of ferrous ammonium sulfate = 0.00139*392 = 0.546 mg ferrous ammonium sulfate................ans 3



mass of ferrous ammonium sulfate = 172mg = 0.172 gm

moles = 0.172/392 = 0.0004387 moles

so moles of Fe = 0.0004387 moles

so mass = 0.0004387*56 = 0.02457 gm

volume = 500ml

so 25.67 mg in 0.5L

so ppm = 25.67/0.5 = 49 ppm....................................ans 4


mass of iron in 1L = 0.049 gm

so mass in 13 ml = 0.049/1000 *13 = 0.000637 gm = 0.637mg

now volume increased to 100 ml


so concentration in ppm = 0.637/0.1= 6.37ppm..............................ans 5

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