100 ml of ethylalcohol is made upon a litre with distilled water. If the density of `C_(2)H_(5)OH` is `0.46gr//ml`. Then its molarity is

Options

A. `0.55 m`
B. `111m`
C. `2.22m`
D. `3.33m`

Explanation

To find the molarity of the ethyl alcohol solution, we will follow these steps:

### Step 1: Calculate the mass of ethyl alcohol
Given the density of ethyl alcohol (C₂H₅OH) is 0.46 g/mL and the volume is 100 mL, we can calculate the mass using the formula:

\[ \text{Mass} = \text{Density} \times \text{Volume} \]

Substituting the values:

\[ \text{Mass} = 0.46 \, \text{g/mL} \times 100 \, \text{mL} = 46 \, \text{g} \]

### Step 2: Calculate the number of moles of ethyl alcohol
Next, we need to calculate the number of moles of ethyl alcohol using its molar mass. The molar mass of C₂H₅OH is 46 g/mol. The number of moles (n) can be calculated using the formula:

\[ n = \frac{\text{Mass}}{\text{Molar Mass}} \]

Substituting the values:

\[ n = \frac{46 \, \text{g}}{46 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 3: Calculate the mass of the solvent (water)
The total volume of the solution is 1 L (1000 mL). Since we have added 100 mL of ethyl alcohol, the volume of distilled water added is:

\[ \text{Volume of water} = 1000 \, \text{mL} - 100 \, \text{mL} = 900 \, \text{mL} \]

The density of water is approximately 1 g/mL, so the mass of the water can be calculated as:

\[ \text{Mass of water} = \text{Density} \times \text{Volume} = 1 \, \text{g/mL} \times 900 \, \text{mL} = 900 \, \text{g} \]

### Step 4: Convert mass of water to kg
To use the mass in kg for molality calculation, we convert grams to kilograms:

\[ \text{Mass of water in kg} = \frac{900 \, \text{g}}{1000} = 0.9 \, \text{kg} \]

### Step 5: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Using the values we have:

\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{1 \, \text{mol}}{0.9 \, \text{kg}} \approx 1.111 \, \text{mol/kg} \]

### Step 6: Final result
Thus, the molality of the solution is approximately:

\[ \text{Molarity} \approx 1.111 \, \text{mol/kg} \]