An antifreeze mixture consists of `40%` ethylene glycol `(C_(2)H_(6)O_(2))` by weight in aqueous solution. If the density of this solution is `1.05 g//"mol"`, what is the molar concentration :

Options

A. `6.77 M`
B. `6.45 M`
C. `0.017 M`
D. `16.9 M`

Explanation

To find the molar concentration (molarity) of the antifreeze mixture consisting of 40% ethylene glycol (C₂H₆O₂) by weight in an aqueous solution, we can follow these steps:

### Step 1: Understand the given data
- Percentage by weight of ethylene glycol (C₂H₆O₂) = 40%
- Density of the solution = 1.05 g/mL
- Molar mass of ethylene glycol (C₂H₆O₂) = 62 g/mol

### Step 2: Calculate the mass of the solution
Assume we have 100 g of the solution for simplicity:
- Mass of ethylene glycol = 40% of 100 g = 40 g
- Mass of water = 100 g - 40 g = 60 g

### Step 3: Calculate the volume of the solution
Using the density of the solution:
\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \]
Rearranging gives us:
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{100 \text{ g}}{1.05 \text{ g/mL}} \approx 95.24 \text{ mL} \]

### Step 4: Convert volume from mL to L
\[ \text{Volume in L} = \frac{95.24 \text{ mL}}{1000} \approx 0.09524 \text{ L} \]

### Step 5: Calculate the number of moles of ethylene glycol
Using the molar mass:
\[ \text{Moles of ethylene glycol} = \frac{\text{mass}}{\text{molar mass}} = \frac{40 \text{ g}}{62 \text{ g/mol}} \approx 0.645 \text{ moles} \]

### Step 6: Calculate the molarity (M)
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.645 \text{ moles}}{0.09524 \text{ L}} \approx 6.77 \text{ M} \]

### Final Answer
The molar concentration of the antifreeze mixture is approximately **6.77 M**.

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