Question

If the density of lake water is 1.092g/ml and 92 g of $Na^+$ ion is present per kg of water. Calculate molarity of $Na^+$ ion in the lake. (Given atomic mass of Na is 23 u)

Answer 1

4 M

Answer 2

Solution:

The given information is:

• Density of lake water (solution) = 1.092 g/ml
• Mass of $Na^+$ ions = 92 g
• Mass of water (solvent) = 1 kg = 1000 g
• Atomic mass of Na = 23 u (Molar mass of $Na^+$ ≈ 23 g/mol)

Step 1: Calculate the number of moles of $Na^+$ ions.

Number of moles = (Mass of $Na^+$) / (Molar mass of $Na^+$)

Number of moles = 92 g / 23 g/mol = 4 moles

Step 2: Calculate the total mass of the solution.

Mass of solution = Mass of $Na^+$ ions + Mass of water

Mass of solution = 92 g + 1000 g = 1092 g

Step 3: Calculate the volume of the solution using the density.

Volume of solution = (Mass of solution) / (Density of solution)

Volume of solution = 1092 g / 1.092 g/ml = 1000 ml

Step 4: Convert the volume of the solution to liters.

Volume of solution in Liters = 1000 ml / 1000 ml/L = 1 L

Step 5: Calculate the molarity of $Na^+$ ions.

Molarity (M) = (Number of moles of $Na^+$) / (Volume of solution in Liters)

Molarity = 4 moles / 1 L = 4 M

Explanation:

1. Calculate moles of $Na^+$ using its mass and molar mass.
2. Find the total mass of the solution by adding the mass of $Na^+$ and the mass of water.
3. Use the density of the lake water (solution) to find the volume of the solution.
4. Convert the volume to liters.
5. Calculate molarity by dividing the moles of $Na^+$ by the volume of the solution in liters.