For an arbitrary nonelectrolyte, calculate the freezing point when $0.0192 mols$ $"0.0192 mols"$ of it is dissolved in $5.00 g$ $"5.00 g"$ of water? $K_{f} = {1.86}^{\circ} C/m$ $K_f = 1.86^@ "C/m"$ for water.

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For an arbitrary nonelectrolyte, calculate the freezing point when $0.0192 mols$ $"0.0192 mols"$ of it is dissolved in $5.00 g$ $"5.00 g"$ of water? $K_{f} = {1.86}^{\circ} C/m$ $K_f = 1.86^@ "C/m"$ for water.

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Answer 1 · 2017-01-26T07:12:45

$- {7.14}^{\circ} C$ $-7.14^@ "C"$ .

The idea is that dissolving anything into a solvent will decrease its freezing point. It is expected that you already know $T_{f}^{*}$ $T_f^"*"$ for water is $0^{\circ} C$ $0^@ "C"$ , i.e. the freezing point of the pure solvent water. Therefore, you are solving for the final, lowered freezing point.

Recall:

$DeltaT_f = T_f - T_f^"*" = -iK_fm$

where:

$i = 1$ is the van't Hoff factor for a non-electrolyte. It simply says that only one nonelectrolyte particle exists for every nonelectrolyte particle that is placed into solution.

$K_f = 1.86^@ "C/m"$ is the freezing point depression constant for water.

$m$ is the molality of the solution, i.e. $"mol solute"/"kg solvent"$ .

$T_f$ is the freezing point of the solution.

$T_f^"*"$ is the freezing point of the pure solvent.

First, calculate what $DeltaT_f$ would be:

$DeltaT_f = -(1)(1.86^@ "C/m")(("0.0192 mols solute")/(5.00xx10^(-3) "kg H"_2"O"))$

$= -7.14^@ "C"$

Therefore, the final freezing point is:

$color(blue)(T_f) = DeltaT_f + T_f^"*" = -7.14^@ "C" + 0^@ "C" = color(blue)(-7.14^@ "C")$

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For an arbitrary nonelectrolyte, calculate the freezing point when $0.0192 mols$ $"0.0192 mols"$ of it is dissolved in $5.00 g$ $"5.00 g"$ of water? $K_{f} = {1.86}^{\circ} C/m$ $K_f = 1.86^@ "C/m"$ for water.

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For an arbitrary nonelectrolyte, calculate the freezing point when "0.0192 mols"0.0192 mols"0.0192 mols" of it is dissolved in "5.00 g"5.00 g"5.00 g" of water? K_f = 1.86^@ "C/m"Kf=1.86∘C/mK_f = 1.86^@ "C/m" for water.

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For an arbitrary nonelectrolyte, calculate the freezing point when $0.0192 mols$ $"0.0192 mols"$ of it is dissolved in $5.00 g$ $"5.00 g"$ of water? $K_{f} = {1.86}^{\circ} C/m$ $K_f = 1.86^@ "C/m"$ for water.