To determine the concentration [tex]\( M_j \)[/tex] of the stock solution, we can use the dilution formula given by:
[tex]\[ M_j \times V_j = M_f \times V_f \][/tex]
Where:
- [tex]\( M_j \)[/tex] is the initial molarity (concentration) of the stock solution.
- [tex]\( V_j \)[/tex] is the initial volume of the stock solution.
- [tex]\( M_f \)[/tex] is the final molarity (concentration) of the diluted solution.
- [tex]\( V_f \)[/tex] is the final volume of the diluted solution.
Given:
- Final volume [tex]\( V_f = 150.0 \)[/tex] mL
- Final molarity [tex]\( M_f = 1.40 \)[/tex] M
- Initial volume [tex]\( V_j = 35.0 \)[/tex] mL
We need to find the initial molarity [tex]\( M_j \)[/tex].
Using the formula, we have:
[tex]\[ M_j \times 35.0 \text{ mL} = 1.40 \text{ M} \times 150.0 \text{ mL} \][/tex]
Rearranging to solve for [tex]\( M_j \)[/tex], we get:
[tex]\[ M_j = \frac{1.40 \text{ M} \times 150.0 \text{ mL}}{35.0 \text{ mL}} \][/tex]
Calculating the right side:
[tex]\[ M_j = \frac{210.0 \text{ M} \cdot \text{mL}}{35.0 \text{ mL}} \][/tex]
[tex]\[ M_j = 6.00 \text{ M} \][/tex]
Hence, the concentration of the stock solution is [tex]\( 6.00 \)[/tex] M.
Therefore, the answer is:
[tex]\[ 6.00 \text{ M} \][/tex]
