Question: The solubility order for alkali metal fluoride in water is :
$\mathrm{LiF}<\mathrm{NaF}<\mathrm{KF}<\mathrm{RbF}$
$\mathrm{LiF}>\mathrm{NaF}>\mathrm{KF}>\mathrm{RbF}$
$\mathrm{RbF}<\mathrm{KF}<\mathrm{NaF}<\mathrm{LiF}$
$\mathrm{LiF}<\mathrm{RbF}<\mathrm{KF}<\mathrm{NaF}$
Correct Option: 1
Solution: