Question

an aqueous solution of glucose (molar mass = 180.2 g/mol) has a molality of 6.43 m and a density of 1.20 g/ml. what is the molarity of glucose in the solution?

Explanation:

Step1: Define molality basis

Use 1 kg (1000 g) of solvent (water) as reference.
Moles of glucose: $n = 6.43\ \text{mol}$

Step2: Calculate mass of glucose

Mass = moles × molar mass
$m_{\text{glucose}} = 6.43\ \text{mol} \times 180.2\ \text{g/mol} = 1158.686\ \text{g}$

Step3: Find total mass of solution

Total mass = mass of solvent + mass of glucose
$m_{\text{total}} = 1000\ \text{g} + 1158.686\ \text{g} = 2158.686\ \text{g}$

Step4: Calculate volume of solution

Volume = total mass / density
$V = \frac{2158.686\ \text{g}}{1.20\ \text{g/mL}} = 1798.905\ \text{mL} = 1.798905\ \text{L}$

Step5: Compute molarity

Molarity = moles of solute / volume of solution
$M = \frac{6.43\ \text{mol}}{1.798905\ \text{L}}$

Answer:

$3.57\ \text{M}$ (rounded to three significant figures)