Question

Calculate the molaity of a solution that is prepared by dissolving 50.4g sucrose (C12H22O11) in 0.332kg of water

Answer 1

Answer:

0.443 mol/kg.

Explanation:

The formula for molal concentration (b) is

$b = \frac{\text{moles of solute}}{\text{kilograms of solvent}}$

$\text{Moles of solute} = \text{50.4 g sucrose} \times \frac{ \text{1 mol sucrose}}{\text{342.30 g sucrose} } = \text{0.1472 mol sucrose}$

$b = \frac{ \text{0.1472 mol}}{ \text{0.332 kg}} = \text{0.443 mol}\cdot\text{kg}^{-1}$