To find the molar solubility of calcium hydroxide $(\text{Ca(OH)}_2)$ in mmol/L, we will go through the following steps:

1. Determine the molar mass of calcium hydroxide (Ca(OH)₂):

   • Calcium (Ca) has a molar mass of approximately 40.08 g/mol.
   • Oxygen (O) has a molar mass of 16.00 g/mol, and there are two hydroxide ions (OH⁻), each with 1 oxygen and 1 hydrogen.
   • Hydrogen (H) has a molar mass of approximately 1.01 g/mol.

   $\text{Molar mass of Ca(OH)}_2 = 40.08 + 2 \times (16.00 + 1.01) = 40.08 + 2 \times 17.01 = 40.08 + 34.02 = 74.10 \, \text{g/mol}$

2. Calculate the molar solubility in mol/L:

   • We know that 0.185 g of Ca(OH)₂ is soluble in 100 mL of water.
   • Convert this to moles using the molar mass.

   $\text{Molar solubility (mol/L)} = \frac{0.185 \, \text{g}}{74.10 \, \text{g/mol}} \times \frac{1000 \, \text{mL}}{100 \, \text{mL}}$

   Simplifying:

   $= \frac{0.185}{74.10} \times 10 = 0.002496 \, \text{mol/L}$

3. Convert mol/L to mmol/L:

   • Since 1 mol/L = 1000 mmol/L, we convert the result:

   $\text{Molar solubility (mmol/L)} = 0.002496 \times 1000 = 2.496 \, \text{mmol/L}$

Answer:

The molar solubility of calcium hydroxide at 0°C is approximately 2.496 mmol/L.

Would you like more details on any step, or do you have further questions?

Here are five related questions to expand your understanding:

1. How does temperature affect the solubility of calcium hydroxide in water?
2. What is the relationship between molar solubility and the solubility product (Ksp) of a compound?
3. How can the solubility be expressed in molarity if the volume of the solution changes?
4. What is the significance of the solubility product constant for sparingly soluble compounds?
5. How would the solubility of calcium hydroxide change if it were dissolved in an acidic solution?

Tip: When converting between units, carefully track each step to ensure precision and accuracy, especially in chemistry calculations where conversions are frequent.