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 solutions when equal weights of both salts are dissolved in equal volumes of solution, we can follow these steps:

### Step 1: Understand the Concept of Molarity
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula for molarity is:

\[ M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]

### Step 2: Calculate the Number of Moles
The number of moles of a solute can be calculated using the formula:

\[ \text{Number of moles} = \frac{\text{Mass of solute (g)}}{\text{Molar mass of solute (g/mol)}} \]

Given that equal weights of NaCl and KCl are dissolved, let’s denote the mass of each solute as \( W \).

### Step 3: Determine the Molar Masses
- **Molar mass of NaCl**:
- Sodium (Na) = 23 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Therefore, Molar mass of NaCl = \( 23 + 35.5 = 58.5 \, \text{g/mol} \)

- **Molar mass of KCl**:
- Potassium (K) = 39 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Therefore, Molar mass of KCl = \( 39 + 35.5 = 74.5 \, \text{g/mol} \)

### Step 4: Calculate the Number of Moles for Each Salt
- For NaCl:
\[ \text{Number of moles of NaCl} = \frac{W}{58.5} \]

- For KCl:
\[ \text{Number of moles of KCl} = \frac{W}{74.5} \]

### Step 5: Calculate Molarity for Each Solution
Since the volume of the solution is the same for both salts, let’s denote it as \( V \) liters.

- Molarity of NaCl:
\[ M_{\text{NaCl}} = \frac{\frac{W}{58.5}}{V} = \frac{W}{58.5V} \]

- Molarity of KCl:
\[ M_{\text{KCl}} = \frac{\frac{W}{74.5}}{V} = \frac{W}{74.5V} \]

### Step 6: Compare the Molarities
To compare the molarities, we can see that since \( W \) and \( V \) are the same for both solutions, the molarity will depend on the molar masses:

\[ \frac{M_{\text{NaCl}}}{M_{\text{KCl}}} = \frac{W/58.5V}{W/74.5V} = \frac{74.5}{58.5} \]

Since \( 74.5 > 58.5 \), it follows that:

\[ M_{\text{NaCl}} > M_{\text{KCl}} \]

### Conclusion
The molarity of the NaCl solution will be greater than that of the KCl solution.

### Final Answer
**The molarity of NaCl solution is greater than that of KCl solution.**

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