A chemist must prepare 500 ml of 9.00uM aqueous mercury(II) iodide working solution. He'll do this by pouring out some 81.0 umol/L aqueous mercury(II) iodide stock solution into a graduated cylinder and diluting it with distilled water.

Calculate the volume in of the mercury(II) iodide stock solution that the chemist should pour out. Be sure your answer has the correct number of significant digits.

it is being diluted 81/9 times , or 9 times.

that means one part original solution, and 8 parts water.
one part stock= 500/9 ml= ...
eight parts water= 8*500/9 = ...

Good job, but how many significant figures should be recorded in the answer?

To calculate the volume of the mercury(II) iodide stock solution that the chemist should pour out, we can use the following formula:

Volume (V1) x Concentration (C1) = Final Volume (V2) x Final Concentration (C2)

where:
V1 = volume of stock solution
C1 = concentration of stock solution
V2 = final volume of working solution (500 mL)
C2 = final concentration of working solution (9.00 µM or 9.00 x 10^-6 M)

Rearranging the formula, we can solve for V1:

V1 = (V2 x C2) / C1

Substituting the given values:

V1 = (500 mL x 9.00 x 10^-6 M) / (81.0 x 10^-6 M)

Calculating the numerator:

(500 mL x 9.00 x 10^-6 M) = 4.50 x 10^-3 mol

Calculating the denominator:

(81.0 x 10^-6 M) = 8.10 x 10^-5 mol/L

Dividing the numerator by the denominator:

V1 = (4.50 x 10^-3 mol) / (8.10 x 10^-5 mol/L)

V1 = 55.6 mL

Therefore, the chemist should pour out approximately 55.6 mL of the mercury(II) iodide stock solution.

To calculate the volume of the mercury(II) iodide stock solution that the chemist should pour out, we can use the dilution equation:

C1V1 = C2V2

Where:
C1 = concentration of the stock solution
V1 = volume of the stock solution poured out
C2 = concentration of the working solution
V2 = final volume of the working solution

Given:
C1 = 81.0 umol/L
V1 = ?
C2 = 9.00 uM
V2 = 500 mL

First, let's convert the concentrations to a consistent unit. Since both concentrations are given in micromoles per liter (umol/L) and the final volume is given in milliliters (mL), we can leave the units as is.

Now, let's substitute the given values into the equation and solve for V1:

(81.0 umol/L) * V1 = (9.00 uM) * (500 mL)

To simplify the units, we can convert 500 mL to liters by dividing it by 1000:

(81.0 umol/L) * V1 = (9.00 uM) * (0.500 L)

Next, let's solve for V1 by dividing both sides of the equation by 81.0 umol/L:

V1 = (9.00 uM) * (0.500 L) / (81.0 umol/L)

V1 ≈ 0.0556 L

Since the volume needs to be reported in milliliters, we can convert 0.0556 L to mL by multiplying by 1000:

V1 ≈ 55.6 mL

Therefore, the chemist should pour out approximately 55.6 mL of the mercury(II) iodide stock solution.