100ml of ethyl alcohol d=0.92g/ml and 900 ml of water d=1g/ml are mixed to form 1 lit solution. The Molarity and molality of the resulting solution are

• 2M and 2m
• 2M and 2.22m
• 2.2M and 1.1m
• 2M and 1m

To find the molarity and molality of the solution formed by mixing ethyl alcohol and water, we first need to calculate the number of moles of ethyl alcohol and the mass of the solvent (water).

Calculating Moles of Ethyl Alcohol

The density of ethyl alcohol is 0.92 g/ml. For 100 ml, the mass can be calculated as:

• Mass of ethyl alcohol = Volume × Density = 100 ml × 0.92 g/ml = 92 g

The molar mass of ethyl alcohol (C2H5OH) is approximately 46 g/mol. Thus, the number of moles is:

• Moles of ethyl alcohol = Mass / Molar mass = 92 g / 46 g/mol = 2 moles

Finding Molarity

Molarity (M) is defined as the number of moles of solute per liter of solution. Since the total volume of the solution is 1 liter:

• Molarity = Moles of solute / Volume of solution in liters = 2 moles / 1 L = 2 M

Calculating Mass of Water

The density of water is 1 g/ml. For 900 ml, the mass is:

• Mass of water = Volume × Density = 900 ml × 1 g/ml = 900 g

Finding Molality

Molality (m) is defined as the number of moles of solute per kilogram of solvent. The mass of water in kilograms is:

• Mass of water in kg = 900 g / 1000 = 0.9 kg

Now, we can calculate molality:

• Molality = Moles of solute / Mass of solvent in kg = 2 moles / 0.9 kg ≈ 2.22 m

Final Results

The resulting solution has:

• Molarity: 2 M
• Molality: 2.22 m

Therefore, the correct answer is 2 M and 2.22 m.