Question

the percent by mass of methanol (molar mass = 32.04 g/mol) in an aqueous solution is 47.3. what is the molality of the methanol solution?

Explanation:

Step1: Define mass of solution components

Assume total solution mass = 100 g.
Mass of methanol: $m_{\text{methanol}} = 47.3\ \text{g}$
Mass of water: $m_{\text{water}} = 100 - 47.3 = 52.7\ \text{g} = 0.0527\ \text{kg}$

Step2: Calculate moles of methanol

Use molar mass to find moles:
$$n_{\text{methanol}} = \frac{m_{\text{methanol}}}{M_{\text{methanol}}} = \frac{47.3\ \text{g}}{32.04\ \text{g/mol}} \approx 1.476\ \text{mol}$$

Step3: Compute molality of solution

Molality = moles of solute per kg of solvent:
$$\text{Molality} = \frac{n_{\text{methanol}}}{m_{\text{water (kg)}}} = \frac{1.476\ \text{mol}}{0.0527\ \text{kg}}$$

Answer:

$\approx 28.0\ \text{mol/kg}$