# Question 38335

Jun 6, 2015

You'd extract 470 g of ice from this solution at that temperature.

The approch here is to determine how much water remains in this solution after it's cooled to that temperature and ice is formed.

The equation for a solution's freezing point depression is

$\Delta {T}_{f} = {K}_{f} \cdot b$, where

$\Delta {T}_{f}$ - the freezing point depression, defined as the difference between the freezing point of pure water and the freezing point of the solution.
${K}_{f}$ - the cryoscopic constant of water;
$b$ - the molality

Your solution's freezing point depression would be

$\Delta {T}_{f} = {1.86}^{\circ} \text{C" cancel("kg") cancel("mol"^(-1)) * 1 cancel("mol")/cancel("kg") = 1.86^@"C}$

Since your cooling temperature is almost twice as low as the solution's normal freezing point, you can expect that close to 50% of the water is now ice.

Since the molality of the solution was 1 molal at a total mass of 1000 g, you know that you have 1 mole of sucrose present.

The mass of the water will decrease, which implies that the molality of the solution, which is expressed as moles of solute per liters of solution, will increase. This means that you can write

DeltaT_"f NEW" = K_f * n_"sucrose"/(x * 10^(-3)"kg"), where

$x$ - the mass of the water expressed in grams.
${n}_{\text{sucrose}}$ - the number of moles of sucrose, in your case 1.

Solve this equation for $x$ to get

3.534^@cancel("C") = 1.86^@cancel("C") * cancel("mol"^(-1)) * cancel("kg") * (1 cancel("mole"))/(x * 10^(-3)cancel("kg"))

$x = \frac{1.86 \cdot 1}{3.534 \cdot {10}^{- 3}} = \text{526.3 g}$

This is how much water your solution contains at that temperature. Thus, you can extract

${m}_{\text{ice" = m_"solution" - m_"water}}$

${m}_{\text{ice" = 1000 - 526.3 = "473.7 g}}$

I'll leave the answer rounded to two sig figs, despite the fact that you only give one sig fig for the mass of the solution and for its molality.

m_"ice" = color(green)("470 g")#

Indeed, almost half of the initial amount of water is now ice.