If $0.64 g$ $"0.64 g"$ of napthalene ( $M = 128.1705 g/mol$ $M = "128.1705 g/mol"$ ) is dissolved in $100 g$ $"100 g"$ a certain solvent ( $T_{f}^{} = 6^{\circ} C$ $T_f^"" = 6^@ "C"$ ), whose $K_{f} = 20^{\circ} C \cdot kg/mol$ $K_f = 20^@ "C"cdot"kg/mol"$ , what is the new freezing point of the solvent?

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If $0.64 g$ $"0.64 g"$ of napthalene ( $M = 128.1705 g/mol$ $M = "128.1705 g/mol"$ ) is dissolved in $100 g$ $"100 g"$ a certain solvent ( $T_{f}^{} = 6^{\circ} C$ $T_f^"" = 6^@ "C"$ ), whose $K_{f} = 20^{\circ} C \cdot kg/mol$ $K_f = 20^@ "C"cdot"kg/mol"$ , what is the new freezing point of the solvent?

$a)$ $a)$ $4^{\circ} C$ $4^@ "C"$
$b)$ $b)$ $5^{\circ} C$ $5^@ "C"$
$c)$ $c)$ $- 1^{\circ} C$ $-1^@ "C"$
$d)$ $d)$ $7^{\circ} C$ $7^@ "C"$

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Answer 1 · 2017-05-20T21:11:57

$(b)$ $(b)$

That's a high $K_{f}$ $K_f$ ... If it weren't for the low concentration, $T_{f}$ $T_f$ may have gotten down to $- 1^{\circ} C$ $-1^@ "C"$ . What would the molality be if the answer were $(a)$ $(a)$ ?

The change in freezing point is given by:

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

where:

$T_f$ is the freezing point of the solution.

$"*"$ indicates pure solvent.

$i$ is the van't Hoff factor of the solute. $i >= 1$ , and corresponds to the effective number of particles per formula unit of solute.

$K_f$ is the freezing point depression constant of the solvent.

$m$ is the molality of the solution in $"mol solute/kg solvent"$ .

$i = 1$ for nonelectrolytes, as they essentially do not dissociate in solution. The molality is the next thing we can calculate...

$m = (0.64 cancel"g Napthalene" xx "1 mol"/(128.1705 cancel"g Napthalene"))/(100 cancel"g solvent" xx "1 kg"/(1000 cancel"g solvent")$

$=$ $"0.050 mol solute/kg solvent"$

Therefore, the change in freezing point is:

$DeltaT_f = -(1)(20^@ "C"cdot"kg/mol")("0.0499 mol/kg")$

$= -0.9987^@ "C" -> -1^@ "C"$

Thus, the new freezing point is:

$color(blue)(T_f) = DeltaT_f + T_f^"*" = (-0.9987^@ "C") + (6^@ "C")$

$~~ color(blue)(5^@ "C")$

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$a)$ $a)$ $4^{\circ} C$ $4^@ "C"$
$b)$ $b)$ $5^{\circ} C$ $5^@ "C"$
$c)$ $c)$ $- 1^{\circ} C$ $-1^@ "C"$
$d)$ $d)$ $7^{\circ} C$ $7^@ "C"$

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a)a)a) 4^@ "C"4∘C4^@ "C" b)b)b) 5^@ "C"5∘C5^@ "C" c)c)c) -1^@ "C"−1∘C-1^@ "C" d)d)d) 7^@ "C"7∘C7^@ "C"

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$a)$ $a)$ $4^{\circ} C$ $4^@ "C"$
$b)$ $b)$ $5^{\circ} C$ $5^@ "C"$
$c)$ $c)$ $- 1^{\circ} C$ $-1^@ "C"$
$d)$ $d)$ $7^{\circ} C$ $7^@ "C"$