36.5% `HCl` has density has density equal to `1.20 g mL^(-1)`. The molarity `(M)` and molality `(m)`, respectively, are

Options

A. 15.7,15.7
B. 12,12
C. 15.7,12
D. 12,15.7

Explanation

To solve the problem, we need to find the molarity (M) and molality (m) of a 36.5% HCl solution with a density of 1.20 g/mL.

### Step-by-Step Solution:

**Step 1: Calculate the Molarity (M)**

1. **Identify the given values:**
- Percentage by weight of HCl = 36.5%
- Density of the solution = 1.20 g/mL
- Molecular weight of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol

2. **Use the formula for molarity:**
\[ M = \frac{\text{Percentage by weight} \times 10 \times \text{Density}}{\text{Molecular weight}} \]

3. **Substitute the values into the formula:**
\[ M = \frac{36.5 \times 10 \times 1.20}{36.5} \]

4. **Calculate:**
\[ M = 10 \times 1.20 = 12 \, \text{M} \]

**Step 2: Calculate the Molality (m)**

1. **Use the formula for molality:**
\[ m = \frac{\text{Percentage by weight} \times 1000}{\text{Molecular weight} \times (100 - \text{Percentage by weight})} \]

2. **Substitute the values into the formula:**
\[ m = \frac{36.5 \times 1000}{36.5 \times (100 - 36.5)} \]

3. **Calculate the denominator:**
\[ 100 - 36.5 = 63.5 \]

4. **Now substitute back into the equation:**
\[ m = \frac{36.5 \times 1000}{36.5 \times 63.5} \]

5. **Cancel out 36.5:**
\[ m = \frac{1000}{63.5} \]

6. **Calculate:**
\[ m \approx 15.75 \, \text{mol/kg} \quad (\text{rounded to 15.7 mol/kg}) \]

### Final Results:
- Molarity (M) = 12 M
- Molality (m) = 15.7 mol/kg