# What is the molar mass of the solute if #"4.18 g"# of it dissolved in #"36.30 g"# of benzene (#K_f = 5.12^@ "C"cdot"kg/mol"#) generates a solution with a freezing point of #2.70^@ "C"#? The freezing point of benzene is #5.53^@ "C"#.

Once you recognize this is freezing point depression, you can use the following equation:

where:

Simply from the units of molality, we have:

Therefore, we can solve for the molar mass later, as long as we can calculate the molality.

Now, we can solve for the molar mass.

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The molar mass of the solute is 78.12 g/mol.

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To find the molar mass of the solute, we can use the formula:

[ \Delta T_f = K_f \cdot m ]

Where:

- ( \Delta T_f ) is the freezing point depression,
- ( K_f ) is the freezing point depression constant (given as ( 5.12^\circ C \cdot kg/mol )),
- ( m ) is the molality of the solution (moles of solute per kilogram of solvent).

First, we calculate the freezing point depression (( \Delta T_f )): [ \Delta T_f = T_f^{\text{pure solvent}} - T_f^{\text{solution}} ] [ \Delta T_f = 5.53^\circ C - 2.70^\circ C = 2.83^\circ C ]

Next, we convert the mass of benzene into kilograms: [ \text{mass of benzene} = 36.30 , \text{g} = 0.0363 , \text{kg} ]

Then, we calculate the molality (( m )) of the solution: [ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} ]

We use the freezing point depression formula to find the moles of solute: [ \Delta T_f = K_f \cdot m ] [ m = \frac{\Delta T_f}{K_f} ] [ m = \frac{2.83^\circ C}{5.12^\circ C \cdot kg/mol} ] [ m = 0.553 , \text{mol/kg} ]

Now, we can use the molality to find the moles of solute: [ \text{moles of solute} = m \times \text{mass of solvent in kg} ] [ \text{moles of solute} = 0.553 , \text{mol/kg} \times 0.0363 , \text{kg} ] [ \text{moles of solute} = 0.0201 , \text{mol} ]

Finally, we calculate the molar mass of the solute: [ \text{molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}} ] [ \text{molar mass} = \frac{4.18 , \text{g}}{0.0201 , \text{mol}} ] [ \text{molar mass} = 207.96 , \text{g/mol} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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