Determine the molarity (mM) for a solution of Na_3Po4 Prepared by ml of a 0.33 t

Question

Determine the molarity (mM) for a solution of Na_3Po4 Prepared by ml of a 0.33 to 1.00L Ml of a 0.33 M solution of Na_3PO_4 to 1.00 L If 5.00 mL of the 1.0 L solution prepared in question 2 was diluted to 50.0 mL, what would be the conncentration (mM) of Na_3PO_4 in this new solution? The absorbance of a 0.100 mM solution of a d ye in a 1.00-cm cell is 0.982 at 420 nm. Calculate the molar absorptivity constant (M^1cM^-1) for this dye. What would be the absorbance of a 0.0500 mM solution of this dye at 420 nm? Assume a 1.00-cm cell is used for this measurement

Explanation / Answer

Question 2

V1 x N1 = V2 x N2

V1 = 10 ml N1 = 0.33M

V2 = 1000 ml N2 = ?

FInd N2 from above equation

N2 = 10 x 0.33 /1000 = 0.0033 M or 3.3 mM

Final concentration of solution = 3.3 mM

Question 3

Apply the same formula here

V1 x N1 = V2 x N2

V1 = 5 ml N1 = 3.3mM

V2 = 50 ml N2 = ?

FInd N2 from above equation

N2 = 10 x 3.3 /50 = 0.66mM

Final concentration of solution = 0.66 mM

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