How much water must be added to $40\mathrm{.}0g$ of $Mg{l}_2$ to produce a solution that is $35\mathrm{.}0\%$ by mass of $Mg{l}_2?$

$87\mathrm{.}5g$

$54\mathrm{.}0g$

$114g$

$74\mathrm{.}3g$

$61\mathrm{.}9g$