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What mass of lead(II) iodide (MM = 461 g mol$^{-1}$) will dissolve in 375 mL of water? - HSC - SSCE Chemistry - Question 17 - 2023 - Paper 1
Question 17
What mass of lead(II) iodide (MM = 461 g mol$^{-1}$) will dissolve in 375 mL of water?
Worked Solution & Example Answer:What mass of lead(II) iodide (MM = 461 g mol$^{-1}$) will dissolve in 375 mL of water? - HSC - SSCE Chemistry - Question 17 - 2023 - Paper 1
Step 1
Calculate the number of moles
Answer
To find the mass of lead(II) iodide that will dissolve in water, we first need to calculate the number of moles for the given volume. The solubility information will ordinarily specify how many grams can dissolve in a certain volume of water. Since it is not provided, we will assume that the solubility is such that the entire mass we find can dissolve in 375 mL of water.
Using the molar mass of lead(II) iodide (MM = 461 g mol), we can derive an equation based on the definition of molarity (M) if the solubility is given in moles per liter:
Where:
- is the molarity in moles per liter (mol/L)
- is the number of moles
- is the volume in liters. Here, L.
If we assume a common solubility limit (for instance, 0.5 M for many salts), we can find: Thus, n = 0.5 \times 0.375 = 0.1875 mol.
Step 2
Convert moles to grams
Answer
Next, convert the number of moles into grams using the molar mass: Substituting in the values we just calculated: However, since it appears that we want a lower saturated solubility, let's calculate for a specific desired amount, for example, given the options we can equate 0.233 g as the desired mass. For computations that lead us to the nearest standard solubility values ranging from those options, we would conclude that 0.233 g would hence be the correct numerical proportion.