To find the concentration of the mercury(I) chloride (Hg₂Cl₂) solution in mmol/L, we can follow these steps:

Step 1: Convert µmol to mmol

We are given the amount of mercury(I) chloride in µmol (0.00124 µmol). To convert µmol to mmol: $1 \, \mu mol = 10^{-3} \, mmol$ So: $0.00124 \, \mu mol = 0.00124 \times 10^{-3} \, mmol = 1.24 \times 10^{-6} \, mmol$

Step 2: Convert the volume from mL to L

The volume of the solution is given as 400 mL. To convert this to liters: $1 \, L = 1000 \, mL$ Thus: $400 \, mL = \frac{400}{1000} \, L = 0.400 \, L$

Step 3: Calculate the concentration in mmol/L

The concentration $C$ in mmol/L is calculated by dividing the amount of substance in mmol by the volume in liters: $C = \frac{\text{amount of substance in mmol}}{\text{volume in L}}$ Substituting the values: $C = \frac{1.24 \times 10^{-6} \, mmol}{0.400 \, L} = 3.10 \times 10^{-6} \, mmol/L$

Final Answer:

The concentration of the mercury(I) chloride solution is: $\boxed{3.10 \times 10^{-6} \, mmol/L}$

Do you want more details or have any questions?

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Here are 5 related questions:

1. How do you convert molarity to other concentration units?
2. What is the significance of significant figures in concentration calculations?
3. Can you explain how to prepare a solution of a given concentration from a stock solution?
4. What is the molecular weight of Hg₂Cl₂ and how would you calculate moles from grams?
5. How would dilution affect the concentration of a solution?

Tip: Always ensure that your units are consistent when performing calculations, especially when converting between µmol, mmol, and mol.