Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Let's Review: The Tools of Quantitative Chemistry - Study Questions - Page 43f: 65


*graph is attached Equation of line => $y= 248x+0.002$ At Absorbance=$0.635$ Concentration = $2.55*10^{-3}$$\frac{g}{L}$ = $2.55*10^{-3}$$\frac{mg}{mL}$

Work Step by Step

To get equation of line: Input the data into a Linear regression which will give you the equation of the line To get Concentration at Absorbance of $0.635$: The equation of the line is for a Absorbance vs Concentration $(g/L)$ Therefore we can input the $0.635$ into the equation of the line as $y$ and then we can solve for $x$ So... $0.635= 248x+.002$ $x= 2.55*10^{-3}\frac{g}{L}$ To get Concentration in $\frac{mg}{mL}$ $\frac{2.55*10^{-3}g}{L}*\frac{1000mg}{1g}*\frac{1L}{1000mL}=2.55*10^{-3}\frac{mg}{mL}$ Thus, Concentration = $2.55*10^{-3}$$\frac{g}{L}$ = $2.55*10^{-3}$$\frac{mg}{mL}$
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