Question

$1000gm$ of $1(m)$ sucrose solution in water is cooled to $- 3.554$ degree centigrade. What mass of ice would be separated out in the nearest possible integers in $gm$ at this temperature? For water ${K_f} = 1.86Kmo{l^{ - 1}}Kg$
(A) $353gm$
(B) $300gm$
(C) $400gm$
(D) $35gm$

Answer 1

Hint : Here as water is being cooled so depression in freezing point will be considered which refers to the lowering of the freezing point of solvents when solutes are added into it. It is a colligative property of the solutions which is directly proportional to the molality of the added solute. The formula used is
$\Delta {T_f} = i \times {K_f} \times m$
Where, $\Delta {K_f} =$ depression in freezing point
$i =$ Van’t Hoff factor
${K_f} =$ cryoscopic constant
$m =$ molality

Complete Step By Step Answer:
We know that depression in freezing point $(\Delta {K_f})$ $=$ freezing point of solvent $-$ freezing point of solution.
$\Delta {K_f} =$ $0 - ( - 3.554)$
So, $\Delta {K_f} =$ $3.534^\circ C$
And ${K_f} = 1.86Kmo{l^{ - 1}}Kg$
We know that one molal of sucrose solution will contain $342$ gm of sucrose per $1000gm$ of water.
So, the weight of solution will be $= 342 \times 1000$
$= 1324gm$
Hence if $1342gm$ contains $1000gm$ of water then,
$1000gm$ of water contains $= \dfrac{{342}}{{1342}} \times 1000$ $gm$ of sucrose
$= 254.85gm$ of sucrose
So, the mass of water in $1000gm$ of water $= 1000 - 254.85$
$= 745.15gm$
Now $\Delta {T_f} = \dfrac{{{K_f} \times W{t_{sucrose}} \times 1000}}{{Mol.W{t_{sucrose}} \times W{t_{water}}}}$
Here the weight of water will be taken as the amount of water left.
So, $\Delta {K_f} =$ $3.534^\circ C$
${K_f} = 1.86Kmo{l^{ - 1}}Kg$
$W{t_{sucrose}} = 254.85gm$
$Mol.W{t_{sucrose}} = 342gm$
Let $W{t_{water}} = x$
On putting the above values in the formula, we get:
$3.534 = \dfrac{{1.86 \times 254.85 \times 1000}}{{342 \times x}}$
So $x = 392.2gm$
Therefore, the mass of ice which is separated out is equal to:
(amount of water in solution $-$ water left after melting)
Hence mass of ice:
$\begin{gathered} = 754.15 - 392.2 \\\ = 352.95gm \\\ \end{gathered}$

Note :
We know that the molecular formula of sucrose is ${C_{12}}{H_{22}}{O_{11}}$ so we calculate the molecular weight of sucrose by simply adding the atomic weight of each element present. Also we know that molality is defined as the number of moles of solute per kilogram of the solvent and the number of moles is equal to the weight divided by the molecular weight, so that is why we expand the formula of depression in freezing point from $\Delta {T_f} = {K_f}m$ to $\Delta {T_f} = \dfrac{{{K_f} \times W{t_{sucrose}} \times 1000}}{{Mol.W{t_{sucrose}} \times W{t_{water}}}}$