We are given the following equation and values:

$MW = K_f \times \left( \frac{m_{\text{solute}}}{m_{\text{solvent}} \times \Delta T_f} \right)$

Where:

• $K_f = 4.49 \, \text{°C} \cdot \text{kg/mol}$
• $m_{\text{solute}} = 0.99 \, \text{g}$
• $m_{\text{solvent}} = 4.85 \, \text{g}$
• $\Delta T_f = 4.7 \, \text{°C}$

To solve for the molecular weight $MW$, we can substitute the given values into the equation. However, we first need to convert the mass of the solvent ($m_{\text{solvent}}$) from grams to kilograms (since $K_f$ has units of °C·kg/mol):

$m_{\text{solvent}} = 4.85 \, \text{g} = 0.00485 \, \text{kg}$

Now, substitute the values into the equation:

$MW = 4.49 \times \left( \frac{0.99}{0.00485 \times 4.7} \right)$

We can simplify the expression step by step:

$MW = 4.49 \times \left( \frac{0.99}{0.022795} \right)$

Now, calculate the fraction inside the parentheses:

$MW = 4.49 \times 43.46$

$MW \approx 195.03 \, \text{g/mol}$

So, the molecular weight $MW$ is approximately 195.03 g/mol.

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Would you like a more detailed explanation of the steps? Or do you have any other questions?

Here are 5 related questions for further exploration:

1. How do you calculate the molality of a solution from mass and moles of solute and solvent?
2. What is the concept of freezing point depression, and how is it related to colligative properties?
3. How do you find the molecular weight of a substance using freezing point depression experiments?
4. How do you convert grams to moles, and why is it important in such calculations?
5. What is the significance of the constant $K_f$ in the equation?

Tip: Always make sure the units of mass are consistent with the unit system used for the freezing point depression constant $K_f$.