Calculate Molarity of a 63% W/W `HNO_(3)` solution if density is
1.4 g/mL :

Options

A. 14M
B. 12M
C. 10M
D. 8M

Explanation

To calculate the molarity of a 63% W/W HNO₃ solution with a density of 1.4 g/mL, we can follow these steps:

### Step 1: Understand the given data
- Weight percentage of HNO₃ (solute) = 63% W/W
- Density of the solution = 1.4 g/mL

### Step 2: Calculate the mass of the solution
Assume we have 100 g of the solution. This makes calculations easier because the weight percentage is given as 63%.

- Mass of the solution = 100 g
- Mass of HNO₃ (solute) = 63% of 100 g = 63 g
- Mass of water (solvent) = 100 g - 63 g = 37 g

### Step 3: Calculate the volume of the solution
Using the density, we can find the volume of the solution.

- Density = Mass/Volume
- Volume = Mass/Density
- Volume of the solution = 100 g / 1.4 g/mL = 71.43 mL

### Step 4: Convert the volume from mL to L
To convert mL to L, we divide by 1000.

- Volume in liters = 71.43 mL / 1000 = 0.07143 L

### Step 5: Calculate the number of moles of HNO₃
To find the number of moles, we need the molar mass of HNO₃.

- Molar mass of HNO₃ = 1 (H) + 14 (N) + 16×3 (O) = 1 + 14 + 48 = 63 g/mol
- Moles of HNO₃ = Mass of HNO₃ / Molar mass of HNO₃ = 63 g / 63 g/mol = 1 mol

### Step 6: Calculate the molarity
Molarity (M) is defined as the number of moles of solute per liter of solution.

- Molarity = Moles of solute / Volume of solution in L
- Molarity = 1 mol / 0.07143 L ≈ 14 M

### Final Answer
The molarity of the 63% W/W HNO₃ solution is approximately **14 M**.

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