Question

How many grams of CH3OH must be added to water to prepare 175 mL of a solution that is 1.00 M CH3OH?

Answer 1

E

Step-by-step 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):

`\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}`

Convert from moles of CH₃OH to grams:

`\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}`

In conclusion, the answer is E.