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In a saline solution that is 5.750% (m/v) potassium iodide $(\ce{KI}),$ there are:

a) 5.750 mL $\ce{KI}$ per 100.0 mL $\ce{H2O}$
b) 575.0 mg $\ce{KI}$ per 10.00 mL of $\ce{KI}$ solution
c) 5.750 g $\ce{KI}$ per 100.0 g $\ce{H2O}$
d) 5.750 g $\ce{KI}$ per 100.0 g of $\ce{KI}$ solution
e) 5.750 g $\ce{KI}$ per 100.0 mL of $\ce{H2O}$

I figured it should be e because for the equation of $(\text{m/v})\%$ is

$$(\text{m/v})\% = \frac{\text{mass solute (g)}}{\text{volume solution (ml)}}$$ and it's $\pu{g/100 mL},$ but it's not the correct answer.

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    $\begingroup$ But 100 mL of ........ ? $\endgroup$ – Poutnik Nov 5 at 6:21
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    $\begingroup$ @Alchimista b looks fine to me without any approximations, why do you think there is no correct answer? An exercise looks decent to me and trains attentiveness well. $\endgroup$ – andselisk Nov 5 at 8:38
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    $\begingroup$ The fact is that writing "there are" means there are. Of course I understand what you mean, but we never write down = instead of ~ . Moreover the use of four digits is disturbing in this context. The solution isn't even in the ppm range, a case that I would have found less or no disturbing. ... $\endgroup$ – Alchimista Nov 5 at 8:44
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    $\begingroup$ @Alchimista I suspect "there are $x$ grams", that's what they mean. I also don't see the problem with four digits, there is no mention how the mass has been determined (and it's really irrelevant for answering). $\endgroup$ – andselisk Nov 5 at 8:50
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    $\begingroup$ The task and the choice from given asnwers is really trivial. I do not understand the big deal @Alchemista makes about it. The author has made a trivial trick with potentially misleading formulation of the correct answer. $\endgroup$ – Poutnik Nov 5 at 8:54
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Since the question raised a discussion in the comments, I guess it won't hurt putting up a brief summarizing answer.

Notation % m/v, as you might've guessed, refers to the ratio between the mass and the volume, or, more precisely, mass concentration $ρ_i$ (sometimes denoted as $γ_i$ [1, p. 48] to distinguish from density, see How to distinguish mass concentration and density?):

$$ρ_i = \frac{m_i}{V}$$

where $m_i$ is the mass of solute and $V$ is the volume of solution, suggesting the SI unit $\pu{kg m-3}.$ Where does percent in % m/v come from? Quoting Wikipedia:

In biology, the "%" symbol is sometimes incorrectly used to denote mass concentration, also called "mass/volume percentage." A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). The notation is mathematically flawed because the unit "%" can only be used for dimensionless quantities.

So, in order to account for this percentage “unit”, one should normalize the mass concentration to the mass of solute in grams in 100 mL of solution. This can be done in one's head in most cases, but you must watch out for the units.

In this particular example, only option b satisfy the definition of the % m/v quantity we are expressing since only b lists volume of KI solution. Let's check the math. Since $\pu{575.0 mg} = \pu{0.5750 g},$ there is $\pu{0.5750 g}$ $\ce{KI}$ per $\pu{10.00 mL}$ of $\ce{KI}$ solution, and normalizing to $\pu{100 mL}$ we indeed obtain the aforementioned concentration:

$$ρ_\ce{KI} = \frac{\pu{0.5750 g}}{\pu{10.00 mL}} = \frac{\pu{5.750 g}}{\pu{100.0 mL}} = 5.750\%~(\text{m/v})$$

The exercise really just requires to pay more attention and follow the analytical chemistry definitions.

References

  1. IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
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