Question #410dd

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Question #410dd

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Answer 1 · 2017-02-24T01:31:32

A) $2$ $2$

Explanation:

The van't Hoff factor, $i$ $i$ , tells you the ratio that exists between the number of moles of solute dissolved in water and the number of moles of particles of solute produced in solution.

$i = \frac{what you get in solution}{what you dissolve to make the solution}$ $i = "what you get in solution"/"what you dissolve to make the solution"$

Potassium hydroxide, $KOH$ $"KOH"$ , is a soluble ionic compound, which means that it dissociates completely to form potassium cations and hydroxide anions

${KOH}_{(a q)} \to K_{(a q)}^{+} + {OH}_{(a q)}^{-}$ $"KOH"_ ((aq)) -> "K"_ ((aq))^(+) + "OH"_ ((aq))^(-)$

Notice that every $1$ $1$ mole of potassium hydroxide that dissolves in solution produces

one mole of potassium cations, $1 \times K^{+}$ $1 xx "K"^(+)$

one mole of hydroxide anions, $1 \times {OH}^{-}$ $1 xx "OH"^(-)$

This means that for every mole of potassium hydroxide you dissolve in solution, you get $2$ $2$ moles of particles of solute.

Therefore, the van't Hoff factor will be equal to $2$ $2$

$i = (1 color(red)(cancel(color(black)("mole K"^(+)))) + 1 color(red)(cancel(color(black)("mole OH"^(-)))))/(1color(red)(cancel(color(black)("mole KOH")))) = 2$

You can thus say that

$color(darkgreen)(ul(color(black)(i = 2))) ->$ you get twice as many moles of particles of solute in solution than the number of moles of solute you dissolved to make the solution

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Question #410dd

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