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°C, we can use the formula for freezing point depression:

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

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

### Step 1: Identify the given values
- \(\Delta T_f = 10°C\)
- \(K_f = 1.86°C \, kg/mol\)
- Mass of water (\(W_a\)) = 600 g = 0.6 kg

### Step 2: Calculate the molality (m)
Using the formula for freezing point depression:

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

Rearranging the formula to find molality \(m\):

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

Substituting the known values:

\[ m = \frac{10°C}{1.86°C \, kg/mol} \approx 5.3763 \, mol/kg \]

### Step 3: Calculate the number of moles of glycerol needed
The molality \(m\) is defined as the number of moles of solute per kilogram of solvent. Therefore, we can calculate the number of moles of glycerol (\(n\)) needed:

\[ n = m \cdot W_a \]

Substituting the values:

\[ n = 5.3763 \, mol/kg \cdot 0.6 \, kg \approx 3.226 \, mol \]

### Step 4: Calculate the mass of glycerol needed
To find the mass of glycerol (\(W_b\)), we use the molar mass of glycerol. The molar mass of glycerol (C₃H₈O₃) is approximately 92 g/mol.

Using the formula:

\[ W_b = n \cdot M \]

Where \(M\) is the molar mass of glycerol. Substituting the values:

\[ W_b = 3.226 \, mol \cdot 92 \, g/mol \approx 296.8 \, g \]

### Final Answer
The weight of glycerol that should be added to 600 g of water to lower its freezing point by 10°C is approximately **296.8 g**.

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