To prepare a 0.40 M NaCl solution starting with 100 mL of a 0.30 M NaCl solution, we need to determine how much NaCl to add to achieve the desired molarity. Here’s a step-by-step solution:
### Step 1: Calculate the moles of NaCl in the initial solution
The initial molarity (M1) is 0.30 M and the volume (V1) is 100 mL (which is 0.100 L).
Using the formula:
\[
\text{Moles of NaCl} = \text{Molarity} \times \text{Volume (in L)}
\]
\[
\text{Moles of NaCl} = 0.30 \, \text{mol/L} \times 0.100 \, \text{L} = 0.030 \, \text{moles}
\]
### Step 2: Determine the final volume needed for 0.40 M
Let’s denote the final molarity (M2) as 0.40 M. We need to find the final volume (V2) that will give us this molarity with the same number of moles of NaCl.
Using the formula:
\[
\text{Moles of NaCl} = \text{Molarity} \times \text{Volume (in L)}
\]
We rearrange to find the final volume:
\[
V2 = \frac{\text{Moles of NaCl}}{\text{Molarity}} = \frac{0.030 \, \text{moles}}{0.40 \, \text{mol/L}} = 0.075 \, \text{L} = 75 \, \text{mL}
\]
### Step 3: Calculate the amount of NaCl to add
Since we want the final volume to be 75 mL but we started with 100 mL of a 0.30 M solution, we need to add NaCl to increase the concentration.
To find out how much more NaCl we need to add to reach 0.40 M, we can calculate the required moles for the final volume of 100 mL (0.100 L):
\[
\text{Required moles for 0.40 M} = 0.40 \, \text{mol/L} \times 0.100 \, \text{L} = 0.040 \, \text{moles}
\]
### Step 4: Calculate the additional moles needed
Now, we subtract the initial moles from the required moles:
\[
\text{Additional moles needed} = 0.040 \, \text{moles} - 0.030 \, \text{moles} = 0.010 \, \text{moles}
\]
### Step 5: Convert moles to grams
Using the molar mass of NaCl (58.5 g/mol):
\[
\text{Mass of NaCl} = \text{Moles} \times \text{Molar Mass} = 0.010 \, \text{moles} \times 58.5 \, \text{g/mol} = 0.585 \, \text{grams}
\]
### Conclusion
To prepare a 0.40 M NaCl solution starting with 100 mL of 0.30 M NaCl, you need to add **0.585 grams of NaCl**.
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