Use the trend line equation from Mg calibration curve to determine the concentra

Question

Use the trend line equation from Mg calibration curve to determine the concentration (ppm) of Mg in each analysis in the dilute multivitamin unknown. Using the dilution and mass data for the multivitamin sample. 0.5234 dissolved in 20 mL of 2M HN03, 2] Filtered 2x w/ 20mL of 0.1 M HN03, diluted to volume w / 0.1 M HN03 in 500 mL volumetric flask, Then diluted 1/10 rightarrow 25 mL volumetric flask. determine the mass in mg for Mg in the multivitamin for each analysis,. calculate the mass in mg for Mg in the multivitamin tablet Mass of tablet = 1.8123g Half of tablet mass used for experiment = 0.5234g

Explanation / Answer

For Multivitamin unknown 1, absorbance = 1.2312

Feed this as y in the trendline equation found by graph with adjusted R^2 value shown above,

y = 0.2467x + 0.2012

1.2312 = 0.2467x + 0.2012

a) concentration of Mg in diluted sample solution x = 3.96 ppm = 3.96 mg

This is present in 250 ml solution

in 25 ml Mg would be = 39.6 mg

b) In original 500 ml (tablet) concentration of Mg would be = 20 x 39.6 = 792 mg

For Multivitamin unknown 1-2, absorbance = 1.2285

Feed this as y in the trendline equation found by graph with adjusted R^2 value shown above,

y = 0.2467x + 0.2012

1.2285 = 0.2467x + 0.2012

a) concentration of Mg in diluted sample solution x = 3.95 ppm = 3.95 mg

This is present in 250 ml solution

in 25 ml Mg would be = 39.5 mg

b) In original 500 ml (tablet) concentration of Mg would be = 20 x 39.5 = 790 mg

For Multivitamin unknown 1-3, absorbance = 1.2244

Feed this as y in the trendline equation found by graph with adjusted R^2 value shown above,

y = 0.2467x + 0.2012

1.2244 = 0.2467x + 0.2012

a) concentration of Mg in diluted sample solution x = 3.93 ppm = 3.93 mg

This is present in 250 ml solution

in 25 ml Mg would be = 39.3 mg

b) In original 500 ml (tablet) concentration of Mg would be = 20 x 39.3 = 786 mg

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