To prepare 0.5M KCI solution from 100 ml of 0.40 M KCI which of the following should be done?

Options

A. add 0.745 g KCI
B. add 20 ml water
C. add 0.1 mole KCI
D. Evaporate 10 ml water

Answer

To prepare a 0.5 M KCl solution from 100 ml of 0.40 M KCl, we need to determine the number of moles of KCl in both the initial and final solutions and then decide on the appropriate action to achieve the desired molarity.

### Step-by-Step Solution:

1. **Calculate the number of moles of KCl in the initial solution:**
- The formula for molarity (M) is given by:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
- Rearranging this gives:
\[ \text{moles of solute} = M \times \text{volume in liters} \]
- For the initial solution:
- Molarity (M) = 0.40 M
- Volume = 100 ml = 0.1 L
\[ \text{moles of KCl} = 0.40 \, \text{M} \times 0.1 \, \text{L} = 0.04 \, \text{moles} \]

2. **Calculate the number of moles of KCl needed for the final solution:**
- For the desired solution:
- Molarity (M) = 0.50 M
- Volume = 100 ml = 0.1 L
\[ \text{moles of KCl needed} = 0.50 \, \text{M} \times 0.1 \, \text{L} = 0.05 \, \text{moles} \]

3. **Determine the additional moles of KCl required:**
- We have 0.04 moles initially and need 0.05 moles.
\[ \text{Additional moles needed} = 0.05 \, \text{moles} - 0.04 \, \text{moles} = 0.01 \, \text{moles} \]

4. **Convert the additional moles of KCl to grams:**
- The molar mass of KCl (Potassium = 39 g/mol, Chlorine = 35.5 g/mol):
\[ \text{Molar mass of KCl} = 39 + 35.5 = 74.5 \, \text{g/mol} \]
- To find the mass required for 0.01 moles:
\[ \text{mass} = \text{moles} \times \text{molar mass} = 0.01 \, \text{moles} \times 74.5 \, \text{g/mol} = 0.745 \, \text{g} \]

5. **Conclusion:**
- To prepare a 0.5 M KCl solution from 100 ml of 0.40 M KCl, you need to **add 0.745 g of KCl** to the solution.