If iodine-131 has a half-life of 8 days, how much of a 64.0 g sample of iodine-131 will remain after 32 day?
1 answer:
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Molar solubility of AgCl will be 0.59 × M.
The amount of a chemical that can dissolve in one liter of a solution before reaching saturation is known as its molar solubility. This implies that the quantity of a substance it can disintegrate in a solution even before the solution becomes saturated with that particular substance is determined by its molar solubility.
A compound's molar solubility would be the measure of how many moles of such a compound must dissolve to produce one liter of saturated solution. The molar solubility unit will be mol L-1.
Calculation of molar solubility:
Given data:
M = 0.30 M
= 1.77 ×
The reaction can be written as:
AgCl ⇔
s s (s+0.30)
= [
]+ [
]
1.77 × = s (0.30)
s = 1.77 × / 0.3
s = 0.59 × M
Therefore, molar solubility of AgCl will be 0.59 × M.
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