Question

A Bill' external standard curve built from absorbance signal vs. analyte concentration (in uM) has the following calibration curve equation y = 0.0500 x + 0.003. If a sample produced an absorbance of 0.300, what was the Bi concentration in the solution in the cuvette? A. 0.015 mM B. 6.00 mm c. 5.94 mm D. 0.600 mm Reset Selection estion 4 of 5 2 Points If the concentration of Bi in a diluted sample (after dilution of the original sample) is found to be 3.00 um, and the original sample was diluted from 0.200 mL to 25.00 mL before analysis, what was the concentration of Bi in the original sample (in mg Bi /30 mL of sample)? A. 1.18 B. 78.4 C. 9.60 D. 2.35 Chat

Answer 1

1) The equation is:
y = 0.05*x + 0.003
Substitute y = 0.300 as observed absorbance and solve for x
x = (y-0.003)/0.05 = (0.3-0.003)/0.05 = 5.94 mM
Thus the answer is option C) 5.94mM.

2)

0.2mL of original solution----> diluted to 25mL and solution B gives concentration = 3.00 $\mu$ M

So, [Bi] in solution B = 3.00*10^-6 mol/L = 3.00*10^-9 mol/mL

Total [Bi] in 25mL = 25mL*3.00*10^-9mol/mL = 75*10^-9 mol

This is also the [Bi] in 0.2mL sample that is diluted.

so the [Bi] in original sample A = 75*10^-9 mol / 0.200mL = 375*10^-9 mol/mL

Volume of the original solution A = 30mL

Hence total [Bi] in original sample solution A = (375*10^-9 mol/mL)*30mL = 11.250*10^-6 mol

Molar mass of Bi = 208.98 g/mol

Mass of Bi in original sample = 208.98g/mol * (11.250*10^-6 mol) = 2351.025 *10^-6 g = 2.35 mg

Thus answer is option D) 2.35 mg