The molarity of` HNO_(3)` in a sample which has density 1.4 g/mL and mass percentage of `63%` is _____.
(Molecular Weight of `HNO_(3) = 63`)

Options

A. 7 M
B. 14 M
C. 2.1 M
D. 28 M

Answer

To find the molarity of HNO₃ in the given sample, we can follow these steps:

### Step 1: Understand the mass percentage
The mass percentage of HNO₃ in the solution is given as 63%. This means that in 100 grams of the solution, there are 63 grams of HNO₃.

### Step 2: Calculate the mass of the solution
Since the mass percentage is based on 100 grams of the solution, we can directly use this value for our calculations.

### Step 3: Use the density to find the volume of the solution
The density of the solution is given as 1.4 g/mL. We can use the formula for density to find the volume of the solution:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Rearranging this gives us:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \text{ g}}{1.4 \text{ g/mL}} \approx 71.428 \text{ mL} \]

### Step 4: Convert volume from mL to L
To convert the volume from mL to L, we divide by 1000:
\[ \text{Volume in L} = \frac{71.428 \text{ mL}}{1000} = 0.071428 \text{ L} \]

### Step 5: Calculate the number of moles of HNO₃
Using the molar mass of HNO₃, which is given as 63 g/mol, we can calculate the number of moles of HNO₃ present in the solution:
\[ \text{Number of moles} = \frac{\text{Mass of solute}}{\text{Molar mass}} = \frac{63 \text{ g}}{63 \text{ g/mol}} = 1 \text{ mol} \]

### Step 6: Calculate the molarity
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{1 \text{ mol}}{0.071428 \text{ L}} \approx 14 \text{ M} \]

### Final Answer
The molarity of HNO₃ in the sample is approximately **14 M**.

---