Websites that end in .com or .org should be avoided.
This problem is providing us with the molality of a solution of calcium iodide as 0.01 m. So the most likely van't Hoff factor is required and theoretically found to be 3 due to the following:
<h3>Van't Hoff factor:</h3>
In chemistry, the correct characterization of solutions also imply the identification of the ions it will release in aqueous solution. For that reason, the van't Hoff factor gives us an idea of this number, according to the formula the solute has got.
In such a way, for calcium iodide, we write its ionization equation as shown below:

Assuming it is able to ionize due to the low molality, because if it was higher, then it won't ionize. Hence, since we have three moles of ion products, one Ca²⁺ and two I⁻, we can conclude the van't Hoff factor would be 3, although calculations may lead to a different, yet close result.
Learn more about the van't Hoff factor: brainly.com/question/23764376
The balanced chemical reaction is:
<span>2C4H10(g)+13O2(g)->10H2O(g)+8CO2(g)
</span>
<span>Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
</span>1.77 g C4H10 (1 mol C4H10/58.14 g C4H10) (10 mol H2O / 2 mol C4H10) ( 1.01 g H2O / 1 mol H2O ) = <span>0.15 g H2O
</span><span>Calculate the mass of butane needed to produce 71.6 of carbon dioxide.
</span>71.6 g CO2 (1 mol CO2/ 44.01 g CO2) ( 2 mol C4H10 / 8 mol CO2 ) (58.14 g C4H10 / 1 mol C4H10 ) = 23.65 g C4H10