Answer:
E
Explanation:
We want to determine the amount of methanol (CH₃OH) needed to prepare 175 mL of a 1.00 M CH₃OH solution.
First, determine the amount of moles of CH₃OH needed. Recalled that molarity is simply moles per liter volume (mol / L):
[tex]\displaystyle\begin{aligned} & 175\text{ mL} \cdot 1.00\text{ M CH$_3$OH} \\ \\ = & 175\text{ mL} \cdot \frac{1.00\text{ mol CH$_3$OH}}{1\text{ L}} \cdot \frac{1 \text{ L}}{1000\text{ mL}} = 0.175\text{ mol CH$_3$OH} \end{aligned}[/tex]
Convert from moles of CH₃OH to grams:
[tex]\displaystyle 0.175\text{ mol CH$_3$OH} \cdot \frac{32.05\text{ g CH$_3$OH}}{1\text{ mol CH$_3$OH}} = 5.61\text{ g CH$_3$OH}[/tex]
In conclusion, the answer is E.
