Question

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

Answer 1

$\Delta T = K _{ f }' \times$ 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 $m'$
$\Delta T = K _{ f }'\times m'$
$m'=\frac{3.534}{1.86}=1.9$
Initially $1000\, gm$ solvent contains $342\, gm$ sucrose
$1342\, gm$ solution contain $342\, gm$ sucrose $1000\, gm$ solution contain
$=\frac{342 \times 1000}{1342}$ sucrose $=254.84$ gm sucrose
Finally,
Amount of water $=1000-254.84$
$=745.16\, gm$
Since sucrose remains same in solution before and after freezing
$\therefore 1.9 \times 342\, gm$ sucrose is in $1000\, gm$ water
$254.84\, gm$ sucrose is in $\frac{1000 \times 254.84}{1.9 \times 342} gm$ $=392.18$ gm water
Thus, wt. of ice separated out
$=745.16-392.18$
$=352.98\, gm$