We’re being asked to calculate for the molality of a 6.35 M methanol (CH_{3}OH) solution.

When calculating for **molality**, we use the following equation:

$\overline{){\mathbf{molality}}{\mathbf{}}\left(\mathbf{m}\right){\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mol}\mathbf{}\mathbf{solute}}{\mathbf{kg}\mathbf{}\mathbf{solvent}}}$

We will calculate the molality of the solution using the following steps:

*Step 1**: Calculate the mass of the solute.*

*Step 2**: Calculate the mass of the solution.*

*Step 3**: Calculate the mass of the solvent (in kg).*

*Step 4**: Calculate the molality of the solution.*

**Solute → CH _{3}OH**

**Solvent → water (H _{2}O)**

Calculate the molality of a 6.35 M aqueous methanol (CH_{3}OH) solution with a density of 0.953 g/mL.

A. 7.28 m

B. 6.99 m

C. 5.87 m

D. 8.47 m