Answer to Question #90149 in General Chemistry for Ida

Question

What is the concetration (mg/L) of 4.5 times 10 to the -4g of resaurin dissolved in distilled water to a volume of 15.0ml, then serially diluted four times using twofold serial dilutions?

Answer 1

Initial concentration of resazurin is

$c=\frac{m}{V}=\frac{4.5\cdot 10^{-4}g}{15.0 mL}=\frac{0.45 mg}{15.0 mL}= 0.03 \frac{mg}{mL}$

After each twofold dilution concentration decreases two times. Thus after four twofold dilutions concentration decreases 2⁴=16 times.

Final concentration of resazurin is $\frac{0.03\frac{mg}{mL}}{16}=0.001875\frac{mg}{mL}=1.875 \times 10^{-3} \frac{mg}{mL}$