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.

d &gt; a &gt; b &gt; c
a &gt; b &gt; c &gt; d
b &gt; c &gt; d &gt; a
c &gt; d &gt; a &gt; b


So, according to what I researched the solubility $s$ for each one of these compounds in water (at $\pu{20 °C})$ is:
$$
\begin{array}{lc}
\hline
\text{Compound} &amp; s/\pu{g L-1}\\
\hline
\ce{NaBr} &amp; 905 \\
\ce{KI} &amp; 1400 \\
\ce{NaF} &amp; 40.4 \\
\ce{NaCl} &amp; 359 \\
\hline
\end{array}
$$
Therefore, my answer would be, from more soluble to least soluble:
b) KI &gt; a) NaBr &gt; d) NaCl &gt; c) NaF
But this answer is not among the list. I think option 4 might be &quot;correct&quot;, 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&gt;b&gt;c&gt;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 ($0.95~ mol L^{-1}$)