Solution, Solute and Solvent Problem 6

Let,
Mass of Solution $= m g$
Mass of Water in this Solution $= x g$
$\Rightarrow$ Mass of $HgCl_2$ in this Solution initially $= (m-x) g$

Using the given Information about solubility,
Solubility in $x g$ of water at $78^oC = 0.25x g$
Solubility in $xg$ of water at $18^oC = 0.06x g$

Initially, at $78^oC$ ,
Mass of $HgCl_2 = m-x = 0.25x$ ------------ 1st equation

After Cooling, at $18^oC$ ,
Mass of Water is still $= xg$
$\Rightarrow$ Mass of $HgCl_2$ finally $= 0.06x$ --------------- 2nd equation

Therefore,
Mass of $HgCl_2$ separated out $=$ 1st equation $-$ 2nd equation
$\Rightarrow 38 = 0.25x-0.06x$ $\Rightarrow x = 200g$

Hence, Using the 1st equation ,
$m = 1.25x = \boxed{250g}$