To determine the molarity of the solution made by the student after adding water, you can use the formula:
\[ M_1V_1 = M_2V_2 \]
Where:
- \( M_1 \) = initial molarity of the NaCl solution (1.50 M)
- \( V_1 \) = initial volume of the NaCl solution (100.5 mL)
- \( M_2 \) = final molarity of the solution after dilution (what we want to find)
- \( V_2 \) = final volume of the solution after dilution (sum of initial volume and added water volume)
Given:
- \( M_1 = 1.50 \, \text{M} \)
- \( V_1 = 100.5 \, \text{mL} \)
- The student added \( 35.5 \, \text{mL} \) of water
First, calculate the final volume after dilution:
\[ V_2 = V_1 + \text{volume of water added} \]
\[ V_2 = 100.5 \, \text{mL} + 35.5 \, \text{mL} = 136.0 \, \text{mL} \]
Now, use the formula to find the final molarity (\( M_2 \)):
\[ M_2 = \frac{M_1 \times V_1}{V_2} \]
\[ M_2 = \frac{(1.50 \, \text{M}) \times (100.5 \, \text{mL})}{136.0 \, \text{mL}} \]
\[ M_2 = \frac{150 \, \text{mol/L} \cdot \text{mL}}{136 \, \text{mL}} \]
\[ M_2 ≈ \frac{150}{136} \, \text{M} \]
\[ M_2 ≈ 1.102 \, \text{M} \]
So, the molarity of the solution made by the student is approximately \( 1.102 \, \text{M} \).
