Order of solubility of sodium halides and potassium iodide


> 
> Order the following ionic compounds in decreasing order according to their ease to dissolve in water.
> 
> 
> **a**) NaBr; **b**) KI; **c**) NaF; **d**) NaCl.
> 
> 
> 1. **d** > **a** > **b** > **c**
> 2. **a** > **b** > **c** > **d**
> 3. **b** > **c** > **d** > **a**
> 4. **c** > **d** > **a** > **b**
> 
> 
> 


So, according to what I researched the solubility $s$ for each one of these compounds in water (at $\pu{20 °C})$ is:


$$
\begin{array}{lc}
\hline
\text{Compound} & s/\pu{g L-1}\\
\hline
\ce{NaBr} & 905 \\
\ce{KI} & 1400 \\
\ce{NaF} & 40.4 \\
\ce{NaCl} & 359 \\
\hline
\end{array}
$$


Therefore, my answer would be, from more soluble to least soluble:


**b**) KI > **a**) NaBr > **d**) NaCl > **c**) NaF


But this answer is not among the list. I think option 4 might be "correct", but it's kind of backwards.


Are the given answers wrong or am I doing something wrong?




The order b-a-d-c is correct, if the solubility is in gramm per liter. If it is in mole per liter, the order is a>b>c>d, as the saturated solutions of NaBr is the highest ($8.78~ mol L^{-1}$), then KI ($8.48~ mol L^{-1}$), then NaCl ($6.19~ mol L^{-1}$), and NaF ($0.95~ mol L^{-1}$)


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