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. 12
B. 14
C. 16
D. 20

Explanation

To find the molarity of HNO₃ in a solution with a density of 1.4 g/mL and a mass percentage of 63%, we can follow these steps:

### Step 1: Calculate the mass of HNO₃ in 1000 g of solution
Given that the mass percentage of HNO₃ is 63%, we can calculate the mass of HNO₃ in 1000 g of solution.

\[ \text{Mass of HNO}_3 = \text{Mass percentage} \times \text{Total mass of solution} \]
\[ \text{Mass of HNO}_3 = 0.63 \times 1000 \, \text{g} = 630 \, \text{g} \]

### Step 2: Calculate the volume of the solution
Using the density of the solution, we can find the volume of 1000 g of solution.

\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
\[ \text{Volume} = \frac{1000 \, \text{g}}{1.4 \, \text{g/mL}} = 714.29 \, \text{mL} \]

### Step 3: Convert the volume from mL to L
To calculate molarity, we need the volume in liters.

\[ \text{Volume in L} = \frac{714.29 \, \text{mL}}{1000} = 0.71429 \, \text{L} \]

### Step 4: Calculate the number of moles of HNO₃
Using the molecular weight of HNO₃ (63 g/mol), we can calculate the number of moles of HNO₃.

\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molecular weight}} = \frac{630 \, \text{g}}{63 \, \text{g/mol}} = 10 \, \text{mol} \]

### Step 5: Calculate the molarity of HNO₃
Molarity (M) is defined as the number of moles of solute per liter of solution.

\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{10 \, \text{mol}}{0.71429 \, \text{L}} \approx 14 \, \text{M} \]

Thus, the molarity of HNO₃ in the solution is approximately **14 M**.

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