Question

a radiator if filled with a mixture of 3.25 kg of ethylene glycol (C2H6O2) in 7.75 jg of water. Calculate the molality of this solution

Answer 1

To calculate the molality of a solution, we use the formula:

\[
\text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}
\]

### Step 1: Calculate the moles of ethylene glycol (C2H6O2)

The molecular weight of ethylene glycol can be calculated as follows:

- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Oxygen (O): 16.00 g/mol

The molecular weight of C₂H₆O₂ (ethylene glycol):

\[
\text{Molar mass} = (2 \times 12.01) + (6 \times 1.01) + (2 \times 16.00)
\]
\[
= 24.02 + 6.06 + 32.00 = 62.08 \text{ g/mol}
\]

Now, we calculate the moles of ethylene glycol in 3.25 kg (which is 3250 g):

\[
\text{Moles of ethylene glycol} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
\]
\[
= \frac{3250 \, \text{g}}{62.08 \, \text{g/mol}} \approx 52.31 \, \text{mol}
\]

### Step 2: Convert the mass of water to kg

The mass of water is given as \( 7.75 \, \text{jg} \). Assuming there's a typo and that it should be \( 7.75 \, \text{kg} \) (since 7.75 jg is extremely small):

\[
\text{Mass of water} = 7.75 \, \text{kg}
\]

### Step 3: Calculate the molality

Now we substitute the values into the molality formula:

\[
\text{Molality (m)} = \frac{52.31 \, \text{mol}}{7.75 \, \text{kg}} \approx 6.73 \, \text{mol/kg}
\]

So, the molality of the solution is approximately **6.73 mol/kg**.