How much grams of `CH_(3)OH` should be dissolved in water for preparing 150 ml of 2.0 M `CH_(3)OH` solution ?

Options

A. `9.6 xx 10^(3)`
B. `2.4 xx 10^(3)`
C. 9.6
D. 2.4

Answer

To solve the problem of how many grams of \( CH_3OH \) (methanol) should be dissolved in water to prepare 150 ml of a 2.0 M \( CH_3OH \) solution, we can follow these steps:

### Step 1: Convert Volume from mL to L
We need to convert the volume of the solution from milliliters to liters because molarity is expressed in moles per liter.

\[ \text{Volume in liters} = \frac{150 \text{ mL}}{1000} = 0.15 \text{ L} \]

### Step 2: Use the Molarity Formula
The molarity (M) of a solution is defined as the number of moles of solute divided by the volume of the solution in liters.

\[ \text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]

Given that the molarity is 2.0 M and the volume is 0.15 L, we can rearrange the formula to find the number of moles of \( CH_3OH \):

\[ \text{Number of moles of } CH_3OH = \text{Molarity} \times \text{Volume in liters} \]

Substituting the values:

\[ \text{Number of moles of } CH_3OH = 2.0 \, \text{M} \times 0.15 \, \text{L} = 0.30 \, \text{moles} \]

### Step 3: Calculate the Mass of \( CH_3OH \)
To find the mass of \( CH_3OH \), we use the formula:

\[ \text{Number of moles} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}} \]

We know the molar mass of \( CH_3OH \) (methanol) is approximately 32 g/mol. Rearranging the formula to find the mass:

\[ \text{Mass (g)} = \text{Number of moles} \times \text{Molar mass (g/mol)} \]

Substituting the values:

\[ \text{Mass (g)} = 0.30 \, \text{moles} \times 32 \, \text{g/mol} = 9.6 \, \text{g} \]

### Conclusion
To prepare 150 ml of a 2.0 M \( CH_3OH \) solution, you need to dissolve **9.6 grams** of \( CH_3OH \) in water.

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