The ratio of number of water molecules in Mohr's salt and potash alum is _________ $$\times 10^{-1}$$. (Integer answer)

First, we recall the chemical formulas of the two hydrated salts involved.

Mohr’s salt is written as $$\text{FeSO}_4\cdot(\text{NH}_4)_2\text{SO}_4\cdot6\text{H}_2\text{O}$$, so it contains $$6$$ molecules of water of crystallisation.

Potash alum (also called potassium alum) is denoted by $$\text{KAl(SO}_4)_2\cdot12\text{H}_2\text{O}$$, which means it contains $$12$$ molecules of water of crystallisation.

We want the ratio of the number of water molecules in Mohr’s salt to those in potash alum. Writing this ratio explicitly, we have

$$ \text{Ratio} \;=\;\frac{\text{Number of } \text{H}_2\text{O molecules in Mohr's salt}}{\text{Number of } \text{H}_2\text{O molecules in potash alum}} \;=\; \frac{6}{12}. $$

Now, dividing the numerator and the denominator by $$6$$ gives

$$ \frac{6}{12}=\frac{1}{2}=0.5. $$

We note that $$0.5$$ can be expressed in scientific notation as

$$ 0.5 \;=\;5\times10^{-1}. $$

Thus, when the ratio is written in the form $$\text{integer}\times10^{-1}$$, the integer required is clearly $$5$$.

Hence, the correct answer is Option 5.