What weight of glycerol should be added to 600 g of water in order to lower its freezing point by `10^@ C` ?
`(K_f=1.86^@ C m^(-1))`

Options

A. 496 g
B. 297 g
C. 310 g
D. 426 g

Explanation

To solve the problem of how much glycerol should be added to 600 g of water to lower its freezing point by \(10^\circ C\), we will use the formula for freezing point depression:

\[ \Delta T_f = K_f \cdot m \]

Where:
- \(\Delta T_f\) is the change in freezing point (in °C),
- \(K_f\) is the freezing point depression constant (in °C kg/mol),
- \(m\) is the molality of the solution (in mol/kg).

### Step 1: Calculate the molality required for the desired freezing point depression.

Given:
- \(\Delta T_f = 10^\circ C\)
- \(K_f = 1.86^\circ C \cdot kg/mol\)

Using the formula:

\[ m = \frac{\Delta T_f}{K_f} = \frac{10}{1.86} \]

Calculating this gives:

\[ m \approx 5.376 \, \text{mol/kg} \]

### Step 2: Calculate the number of moles of glycerol needed.

Next, we need to find out how many moles of glycerol are required to achieve this molality in 600 g of water.

First, convert the mass of water into kg:

\[ 600 \, \text{g} = 0.6 \, \text{kg} \]

Using the definition of molality:

\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \implies \text{moles of solute} = m \cdot \text{kg of solvent} \]

Substituting the values:

\[ \text{moles of glycerol} = 5.376 \, \text{mol/kg} \cdot 0.6 \, \text{kg} \approx 3.226 \, \text{mol} \]

### Step 3: Calculate the mass of glycerol required.

Now, we need to find the mass of glycerol. The molar mass of glycerol (C₃H₈O₃) is calculated as follows:

- Carbon (C): 12.01 g/mol × 3 = 36.03 g/mol
- Hydrogen (H): 1.008 g/mol × 8 = 8.064 g/mol
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol

Adding these together gives:

\[ \text{Molar mass of glycerol} = 36.03 + 8.064 + 48.00 = 92.094 \, \text{g/mol} \]

Now, we can calculate the mass of glycerol needed:

\[ \text{mass of glycerol} = \text{moles of glycerol} \times \text{molar mass of glycerol} \]

Substituting the values:

\[ \text{mass of glycerol} = 3.226 \, \text{mol} \times 92.094 \, \text{g/mol} \approx 297.00 \, \text{g} \]

### Final Answer:
The weight of glycerol that should be added to 600 g of water to lower its freezing point by \(10^\circ C\) is approximately **297 g**.

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