# Question b1235

Jun 7, 2015

You'll remove 39 g of ice from your solution if you coold it to that temperature.

The freezing point of pure water is ${0}^{\circ} \text{C}$. When you mix the ethylene glycol sample with that much water, your solution's freezing point will be lower than ${0}^{\circ} \text{C}$.

You can get an idea about how much ice you'll be able to remove if you compared the freezing point of the solution with the temperature you cool it to.

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.

Use ethylene glycol's malor mass to determine how many moles you have present in solution.

50cancel("g") * "1 mole"/(62.07cancel("g")) = "0.806 moles" ${C}_{2} {H}_{6} {O}_{2}$

The initial solution's molality will be

$b = {n}_{\text{ethylene glycol"/m_"water" = "0.806 moles"/(200 * 10^(-3)"kg") = "4.03 molal}}$

$\Delta {T}_{f} = {1.86}^{\circ} \text{C"cancel("mol"^(-1)) * cancel("kg") * 4.03cancel("mol")/cancel("kg") = 7.50 ^@"C}$

Notice that you're cooling the solution to a temperature that's lower than $\text{-7.50"^@"C}$. This implies that the molality of the solution will increase because some of the water turns to ice, thus leaving less water in the solution.

The number of mole of ethylene glycol will remain unchanged, so you can write

$\Delta {T}_{\text{f cool" = K_f * b_"new}}$, where

${b}_{\text{new}}$ - the molality of the solution at this new temperature.

This will get you, using $x$ to represent the remaing mass of water in grams

9.3^@cancel("C") = 1.86^@cancel("C") cancel("mol"^(-1)) cancel("kg") * (0.806cancel("moles"))/(x * 10^(-3)cancel("kg"))

$9.3 = \frac{1.499 \cdot {10}^{3}}{x} \implies x = \frac{1499}{9.3} = \text{161.18 g}$

This is how much water your solution will contain at $\text{-9.3"^@"C}$. As a result, the amount of ice that can be extracted will be

${m}_{\text{ice" = 200 - 161.18 = "38.82 g}}$

Rounded to two sig figs, although you only gave one sig fig for the mass of water, the answer will be

m_"ice" = color(green)("39 g")#