A mixture containing 41.0 g of an unknown nonelectrolyte and 160.0 g of water has a freezing point of -1.34 °C.Given for water,what is the molecular weight (g/mol)of the unknown solute?
A)57.0
B)356
C)0.720
D)4.74
E)0.355

Question

A mixture containing 41.0 g of an unknown nonelectrolyte and 160.0 g of water has a freezing point of -1.34 °C.Given  for water,what is the molecular weight (g/mol)of the unknown solute?
A)57.0 
B)356 
C)0.720 
D)4.74 
E)0.355

Quizplus · Accepted Answer

The freezing point depression ($\Delta T_f$) is given by: 
$$\Delta T_f = K_f \cdot m$$
where $K_f$ is the freezing point depression constant for water (1.86 °C/m) and $m$ is the molality of the solution (moles of solute per kilogram of solvent).

First, we need to calculate the molality of the solution:

$$m = \frac{n_	ext{solute}}{m_	ext{solvent}} = \frac{	ext{mol}}{	ext{kg}}$$

To find the number of moles of solute, we can use the formula:

$$n_	ext{solute} = \frac{m_	ext{solute}}{M_	ext{solute}}$$

where $M_	ext{solute}$ is the molar mass of the solute.

Substituting the given values:

\begin{align*}
\Delta T_f &= K_f \cdot m \
-1.34 &= (1.86\ \frac{	ext{°C}}{	ext{m}})\cdot m \
m &= -\frac{1.34\ 	ext{°C}}{1.86\ \frac{	ext{°C}}{	ext{m}}} = -0.72\ \frac{	ext{mol}}{	ext{kg}}
\end{align*}

Since the solute is a nonelectrolyte, it does not dissociate into ions and the van't Hoff factor ($i$) is equal to 1. Therefore, we can use the formula:

$$\Delta T_f = K_f \cdot i \cdot m = K_f \cdot m$$

Substituting the values and solving for the number of moles:

\begin{align*}
\Delta T_f &= K_f \cdot m \
-1.34 &= (1.86\ \frac{	ext{°C}}{	ext{m}})\cdot (-0.72\ \frac{	ext{mol}}{	ext{kg}}) \
n_	ext{solute} &= \frac{m_	ext{solute}}{M_	ext{solute}} = \frac{(-1.34\ 	ext{°C})}{(1.86\ \frac{	ext{°C}}{	ext{m}})\cdot(-0.72\ \frac{	ext{mol}}{	ext{kg}})} \
&= 356\ 	ext{g/mol}
\end{align*}

Therefore, the molecular weight of the unknown solute is 356 g/mol.

A Mixture Containing 41