# Question #38335

##### 1 Answer

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

*freezing point depression*, defined as the difference between the freezing point of *pure water* and the freezing point of the solution.

Your solution's freezing point depression would be

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

**1**.

Solve this equation for

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

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.

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