Let's solve this problem step-by-step.
We are given the following information:
- Final volume ([tex]\( V_f \)[/tex]) of the solution is 150.0 mL.
- Final concentration ([tex]\( M_f \)[/tex]) of the solution is 1.40 M.
- Volume of the stock solution ([tex]\( V_i \)[/tex]) is 35.0 mL.
We need to find the concentration of the stock solution ([tex]\( M_i \)[/tex]).
We can use the dilution equation, which is given by:
[tex]\[ M_i V_i = M_f V_f \][/tex]
Substitute the given values into the equation:
[tex]\[ M_i \cdot 35.0 \text{ mL} = 1.40 \text{ M} \cdot 150.0 \text{ mL} \][/tex]
First, calculate the right side of the equation:
[tex]\[ 1.40 \text{ M} \times 150.0 \text{ mL} = 210.0 \text{ M} \cdot \text{mL} \][/tex]
Now, we solve for [tex]\( M_i \)[/tex] by dividing both sides by 35.0 mL:
[tex]\[ M_i = \frac{210.0 \text{ M} \cdot \text{mL}}{35.0 \text{ mL}} \][/tex]
Simplify the division:
[tex]\[ M_i = 6.00 \text{ M} \][/tex]
Therefore, the concentration of the stock solution is [tex]\( 6.00 \text{ M} \)[/tex].
Among the given options, the correct answer is:
[tex]\[ 6.00 \text{ M} \][/tex]
