To find the concentration of the stock solution (denoted as [tex]\( M_i \)[/tex]), we can use the dilution equation:
[tex]\[ M_i \times V_i = M_f \times V_f \][/tex]
where:
- [tex]\( M_i \)[/tex] is the initial molarity (concentration) of the stock solution,
- [tex]\( V_i \)[/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 data:
- [tex]\( M_f = 1.40 \)[/tex] M (final molarity),
- [tex]\( V_f = 150.0 \)[/tex] mL (final volume),
- [tex]\( V_i = 35.0 \)[/tex] mL (initial volume).
We need to find [tex]\( M_i \)[/tex]. Rearrange the dilution equation to solve for [tex]\( M_i \)[/tex]:
[tex]\[ M_i = \frac{M_f \times V_f}{V_i} \][/tex]
Now, substitute the given values into the equation:
[tex]\[ M_i = \frac{1.40 \, \text{M} \times 150.0 \, \text{mL}}{35.0 \, \text{mL}} \][/tex]
Perform the multiplication in the numerator:
[tex]\[ 1.40 \, \text{M} \times 150.0 \, \text{mL} = 210.0 \, \text{M} \cdot \text{mL} \][/tex]
Then, divide by the initial volume:
[tex]\[ M_i = \frac{210.0 \, \text{M} \cdot \text{mL}}{35.0 \, \text{mL}} \][/tex]
[tex]\[ M_i = 6.0 \, \text{M} \][/tex]
So, the concentration of the stock solution is:
[tex]\[ 6.00 \, \text{M} \][/tex]
Therefore, the correct answer is [tex]\( 6.00 \, \text{M} \)[/tex].
