Question

$1000gm$ of $1m$ sucrose solution in water is cooled to $-3.534^{\circ }C$. What weight of ice would be separated out at this temperature ?
$K_f\left(H_{2}O\right)=1.86K$ mole $kg$.

Answer 1

Answer: 352.98

$\Delta T=K_f^{\prime }\times$ molality
$=1.86\times 1=1.86$
$\therefore$ solution starts freezing at $-1.86^{\circ }C$.
Thus, on cooling upto $-3.534^{\circ }C$, freezing continues.
Let molality of solution left at $-3.534$ be $m^{\prime }$
$\Delta T=K_f^{\prime }\times m^{\prime }$
$m^{\prime }=\frac{3.534}{1.86}=1.9$
Initially $1000gm$ solvent contains $342gm$ sucrose
$1342gm$ solution contain $342gm$ sucrose $1000gm$ solution contain
$=\frac{342\times 1000}{1342}$ sucrose $=254.84$ gm sucrose
Finally,
Amount of water $=1000-254.84$
$=745.16gm$
Since sucrose remains same in solution before and after freezing
$\therefore 1.9\times 342gm$ sucrose is in $1000gm$ water
$254.84gm$ sucrose is in $\frac{1000\times 254.84}{1.9\times 342}gm$ $=392.18$ gm water
Thus, wt. of ice separated out
$=745.16-392.18$
$=352.98gm$