Equal weight of NaCl and KCl are dissolved separately in equal volumes of solutions, then the molarity

Options

A. Will be equal for the two solutions
B. For NaCl solution will be greater than that of KCl solution
C. For KCl solution will be greater than that of NaCl solution
D. For NaCl solution will be half of that of KCl solution

Answer

To solve the problem of determining the molarity of NaCl and KCl when equal weights are dissolved in equal volumes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Molar Masses**:
- For NaCl:
- Sodium (Na) = 23 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
- For KCl:
- Potassium (K) = 39 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

2. **Define the Given Weight**:
- Let the equal weight of NaCl and KCl be denoted as \( W \) grams.

3. **Calculate the Number of Moles**:
- For NaCl:
\[ \text{Number of moles of NaCl} = \frac{W}{58.5} \]
- For KCl:
\[ \text{Number of moles of KCl} = \frac{W}{74.5} \]

4. **Define the Volume of Solution**:
- Let the volume of the solution for both NaCl and KCl be \( V \) liters.

5. **Calculate Molarity**:
- Molarity (M) is defined as the number of moles of solute per liter of solution.
- For NaCl:
\[ \text{Molarity of NaCl} = \frac{\text{Number of moles of NaCl}}{V} = \frac{W/58.5}{V} = \frac{W}{58.5 \cdot V} \]
- For KCl:
\[ \text{Molarity of KCl} = \frac{\text{Number of moles of KCl}}{V} = \frac{W/74.5}{V} = \frac{W}{74.5 \cdot V} \]

6. **Compare Molarities**:
- Since both solutions have the same weight \( W \) and volume \( V \), we can compare the molarities:
\[ \text{Molarity of NaCl} = \frac{W}{58.5 \cdot V} \]
\[ \text{Molarity of KCl} = \frac{W}{74.5 \cdot V} \]
- Since \( 58.5 < 74.5 \), it follows that:
\[ \text{Molarity of NaCl} > \text{Molarity of KCl} \]

### Conclusion:
The molarity of the NaCl solution will be greater than that of the KCl solution when equal weights are dissolved in equal volumes.