Answer to Question #86789 in General Chemistry for Lim Yoona

Question #86789

Determination of the new concentration of Dye and NaClO

A student has the following solutions available to them.
- Dye (1.23 x 10-6 mol L-1) – set to dispense in 1 mL aliquots.
- Deionised water – set to dispense in 0.5 mL aliquots
- NaClO (0.8 mol L-1) dispensed using a graduated pipette.

The student is assigned to Run 1. Calculate the initial concentration of dye and bleach in this run. Fill in only the blank part of the answer, to 2 significant figures.

Run 1
Dye (3.0mL)

Deionised water (1.0mL)

NaClO (0.50mL)

New concentration of Dye (mol L-1) ＝ ___ x 10^-7

New concentration of NaClO (mol L-1) ＝ ___ x 10^-2

Expert's answer

New concentration of Dye (mol L-1) ＝ 8.2 x 10^-7

New concentration of NaClO (mol L-1) ＝ 8.88 x 10^-2