Given a solution of `HNO_3` of density 1.4 g/mL and 63% w/w. Determine molarity of `HNO_3` solution.

Options

A. (a) 11 M
B. (b) 12 M
C. (c) 13 M
D. (d) 14 M

Explanation

To determine the molarity of a solution of HNO3 with a density of 1.4 g/mL and a concentration of 63% w/w, we can follow these steps:

### Step 1: Understand the meaning of 63% w/w
63% w/w means that in 100 g of the solution, there are 63 g of HNO3.

### Step 2: Calculate the mass of the solution
Since we are considering 100 g of the solution, the mass of the solution is:
- Mass of solution = 100 g

### Step 3: Calculate the mass of HNO3
From the percentage given, the mass of HNO3 in the solution is:
- Mass of HNO3 = 63 g

### Step 4: Calculate the volume of the solution using density
Density is defined as mass per unit volume. We can rearrange the formula for density to find the volume:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Given that the density of the solution is 1.4 g/mL:
\[ \text{Volume} = \frac{100 \, \text{g}}{1.4 \, \text{g/mL}} = 71.428 \, \text{mL} \]

### Step 5: Calculate the molar mass of HNO3
The molar mass of HNO3 (Hydrogen Nitric Acid) is:
- H: 1 g/mol
- N: 14 g/mol
- O: 16 g/mol × 3 = 48 g/mol
- Total = 1 + 14 + 48 = 63 g/mol

### Step 6: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution. First, we need to convert grams of HNO3 to moles:
\[ \text{Moles of HNO3} = \frac{\text{Mass of HNO3}}{\text{Molar mass of HNO3}} = \frac{63 \, \text{g}}{63 \, \text{g/mol}} = 1 \, \text{mol} \]
Now, convert the volume from mL to L:
\[ \text{Volume in L} = \frac{71.428 \, \text{mL}}{1000} = 0.071428 \, \text{L} \]
Now, calculate the molarity:
\[ \text{Molarity} = \frac{\text{Moles of HNO3}}{\text{Volume in L}} = \frac{1 \, \text{mol}}{0.071428 \, \text{L}} \approx 14 \, \text{M} \]

### Final Answer:
The molarity of the HNO3 solution is approximately **14 M**.

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