# What is the change in the freezing point of water when 35.0 g of sucrose is dissolved in 300.0 g of water?

## Kf of water = -1.86 degC/mol molar mass sucrose = 342.30 g/mol i value of sugar = 1

Jul 31, 2017

$\Delta {T}_{f} = - 0.634$ $\text{^"o""C}$

#### Explanation:

We're asked to find the freezing point depression of a solution.

To do this, we use the equation

DeltaT_f = i·m·K_f

where

• $\Delta {T}_{f}$ is the change in freezing point temperature (what we're trying to find)

• $i$ is the van't Hoff factor, which is given as $1$ (and usually is $1$ in the case of nonelectrolytes)

• $m$ is the molality of the solution, which is

$\text{molality" = "mol solute"/"kg solvent}$

Convert the given mass of sucrose to moles using its molar mass:

35.0cancel("g sucrose")((1color(white)(l)"mol sucrose")/(342.30cancel("g sucrose"))) = color(red)(0.102 color(red)("mol sucrose"

The molality is thus

"molality" = color(red)(0.120color(white)(l)"mol sucrose")/(0.3000color(white)(l)"kg water") = color(green)(0.341m

• ${K}_{f}$ is the molal freezing point constant for the solvent (water), which is given as $- 1.86$ $\text{^"o""C/} m$

Plugging in known values, we have

$\Delta {T}_{f} = \left(1\right) \left(\textcolor{g r e e n}{0.341} \cancel{\textcolor{g r e e n}{m}}\right) \left(- 1.86 \textcolor{w h i t e}{l} \text{^"o""C/} \cancel{m}\right)$

= color(blue)(ul(-0.634color(white)(l)""^"o""C"

This represents by how much the freezing point decreases. The new freezing point of the solution is found by adding this value from the normal freezing point of the solvent ($0.0$ $\text{^"o""C}$ for water):

$\text{new f.p.} = 0.0$ $\text{^"o""C}$ - color(blue)(0.634 color(blue)(""^"o""C" = ul(-0.634color(white)(l)""^"o""C"