# How do you convert 11.0 g CH_3OH to 0.490 M?

Dec 1, 2016

You want to make a methanolic solution in water whose concentration with respect to methanol is $0.490 \cdot m o l \cdot {L}^{-} 1$?

#### Explanation:

And thus $\text{Moles of methanol"/"Volume of solution}$ $=$ $0.490 \cdot m o l \cdot {L}^{-} 1$

$\text{Methanol}$ has a formula mass of $32.04 \cdot g \cdot m o {l}^{-} 1$.

And thus $\frac{\frac{11.0 \cdot g}{32.04 \cdot g \cdot m o {l}^{-} 1}}{\text{Volume of solution}} = 0.490 \cdot m o l \cdot {L}^{-} 1$.

Upon rearrangement, $\text{Volume of solution} = \frac{\frac{11.0 \cdot g}{32.04 \cdot g \cdot m o {l}^{-} 1}}{0.490 \cdot m o l \cdot {L}^{-} 1}$, which is under $1 \cdot L$.

Is this answer dimensionally consistent? Does it give an answer in $L$, which is required?